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25 questions on solving quadratic equations

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The smallest root is $x=$[[0]] and the greatest root is $x=$[[1]]

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solve ax^2+bx+c=0 when a = 1 roots are integers

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$\\simplify{ x ^ 2 + {b} * x + {c }}=0$

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The smallest root is $x=$[[0]] and the greatest root is $x=$[[1]]

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solve ax^2+bx+c=0 when a = 1 roots are integers

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$\\simplify{ x ^ 2 + {b} * x + {c }}=0$

The smallest root is $x=$[[0]] and the greatest root is $x=$[[1]]

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solve ax^2+bx+c=0 when a = 1 roots are integers

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$\\simplify{ x ^ 2 + {b} * x + {c }}=0$

The smallest root is $x=$[[0]] and the greatest root is $x=$[[1]]

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solve ax^2+bx+c=0 when a = 1 roots are integers

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$\\simplify{ x ^ 2 + {b} * x + {c }}=0$

The smallest root is $x=$[[0]] and the greatest root is $x=$[[1]]

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solve ax^2+bx+c=0 when a = 1 roots are integers

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$\\simplify{ x ^ 2 + {b} * x + {c }}=0$

$2x^2 + 13x+20=0$

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-4

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-5

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-2

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0.4

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-2.5

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$x^2-4x+4=0$

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-1

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-2

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-4

$4x^2+x-14=0$

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-2

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-1

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7/4

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4/7

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$4x^2+12x+9=0$

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1.5

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-1.5

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-2

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-2/3

$12x^2-2x-10=0$

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-6/5

Find the solutions of the following quadratics. If a root is repeated then only provide one answer.

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Nothing of interest in advice section

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Exercises in solving quadratics - guess the solutions

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Consider the graph below.

{geogebra_applet('https://ggbm.at/cQSfYQ2k', defs, [])}

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$y=|\\simplify[fractionNumbers,unitFactor]{{a}x+{b}}|$

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$y=\\simplify[fractionNumbers,unitFactor]{{a}x+{b}}$

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$y=|\\simplify[fractionNumbers,unitFactor]{{-a}x+{b}}|$

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$y=\\simplify[fractionNumbers,unitFactor]{{a}abs(x)+{b}}$

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Select the equation for this graph.

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This is not a function. The graph fails the vertical line test, because a vertical line would usually touch two points on the curve.

This is a one-to-many map. That means each input ($x$-axis, domain) often go to two outputs ($y$-axis, range).

This relation does not have an inverse function. Although it passes the horizontal line test....remember, the horizontal line test only applies to functions!

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This uses an embedded Geogebra graph of amodulus function with random coefficients set by NUMBAS.

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Find the roots of the following quadratic equation.

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Solve for $x$: $\\displaystyle ax ^ 2 + bx + c=0$.

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Direct Factorisation

If you can spot a direct factorisation then this is the quickest way to do this question.

For this example we have the factorisation

\\[\\simplify{{a*b} * y ^ 2 + ( {-b*c-a * d}) * y + {c * d} = ({a} * y + { -c}) * ({b} * y + { -d})}\\]

Hence we find the roots:
\\[\\begin{eqnarray} y&=& \\simplify{{n1-n4}/{2*a*b}}\\\\ y&=& \\simplify{{n1+n4}/{2*a*b}} \\end{eqnarray} \\]

Other Methods.

There are several methods of finding the roots – here are the main methods.

Finding the roots of a quadratic using the standard formula.

We can use the following formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

The two roots are

\\[ x = \\frac{-b +\\sqrt{b^2-4ac}}{2a}\\mbox{ and } x = \\frac{-b -\\sqrt{b^2-4ac}}{2a}\\]
there are three main types of solutions depending upon the value of the discriminant $\\Delta=b^2-4ac$

1. $\\Delta \\gt 0$. The roots are real and distinct

2. $\\Delta=0$. The roots are real and equal. Their common value is $\\displaystyle -\\frac{b}{2a}$

3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.

For this question the discriminant of $\\simplify{{a*b}x^2+{-b*c-a*d}x+{c*d}}$ is $\\Delta = \\simplify[std]{{-n1}^2-4*{a*b*c*d}}=\\var{disc}$

{rdis}.

So the {rep} roots are:

\\[\\begin{eqnarray} x = \\frac{\\var{n1} - \\sqrt{\\var{disc}}}{\\var{n3}} &=& \\frac{\\var{n1} - \\var{n4} }{\\var{n3}} &=& \\simplify{{n1 - n4}/ {n3}}\\\\ x = \\frac{\\var{n1} + \\sqrt{\\var{disc}}}{\\var{n3}} &=& \\frac{\\var{n1} + \\var{n4} }{\\var{n3}} &=& \\simplify{{n1 + n4}/ {n3}} \\end{eqnarray}\\]

Completing the square.

First we complete the square for the quadratic expression $\\simplify{{a*b}x^2+{-n1}x+{c*d}}$
\\[\\begin{eqnarray} \\simplify{{a*b}x^2+{-n1}x+{c*d}}&=&\\var{n5}\\left(\\simplify{x^2+({-n1}/{a*b})x+ {c*d}/{a*b}}\\right)\\\\ &=&\\var{n5}\\left(\\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2+ \\simplify{{c*d}/{a*b}-({-n1}/({2*a*b}))^2}\\right)\\\\ &=&\\var{n5}\\left(\\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2 -\\simplify{ {n2^2}/{4*(a*b)^2}}\\right) \\end{eqnarray} \\]
So to solve $\\simplify{{a*b}x^2+{-n1}x+{c*d}}=0$ we have to solve:
\\[\\begin{eqnarray} \\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2&\\phantom{{}}& -\\simplify{ {n2^2}/{4*(a*b)^2}}=0\\Rightarrow\\\\ \\left(\\simplify{x+({-n1}/{2*a*b})}\\right)^2&\\phantom{{}}&=\\simplify{ {n2^2}/{4*(a*b)^2}=({abs(n2)}/{2*a*b})^2} \\end{eqnarray}\\]
So we get the two {rep} solutions:
\\[\\begin{eqnarray} \\simplify{x+({-n1}/{2*a*b})}&=&\\simplify{-{abs(n2)}/{2*a*b}} \\Rightarrow &x& = \\simplify{({-abs(n2)+n1}/{2*a*b})}\\\\ \\simplify{x+({-n1}/{2*a*b})}&=&\\simplify{({abs(n2)}/{2*a*b})} \\Rightarrow &x& = \\simplify{({n1+abs(n2)}/{2*a*b})} \\end{eqnarray}\\]

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Input numbers as fractions or integers not as a decimals.

Solve for $x$: \\[\\simplify[std]{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d}}=0\\]
The least root is $x=\\;$ [[0]]. The greatest root is $x=\\;$ [[1]]

You can get more information on solving a quadratic by clicking on Show steps. You will lose 1 mark if you do so.

Enter the least root first. If the roots are equal, enter the root in both input boxes.

Enter the roots as fractions or integers, not as decimals.

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Finding the roots by factorisation.

Finding a factorisation of a quadratic $q(x)=a(x-r)(x-s)$ where $a$ is the coefficient of $x^2$ gives the roots $x=r$, $x=s$ immediately.

If you cannot find a factorisation then there are several other methods you can use.

Using the formula for the roots.

You can find the roots by using the formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

The two roots are:

\\[ x = \\frac{-b +\\sqrt{b^2-4ac}}{2a}\\mbox{ and } x = \\frac{-b -\\sqrt{b^2-4ac}}{2a}\\]
there are three main types of solutions depending upon the value of the discriminant $\\Delta=b^2-4ac$

1. $\\Delta \\gt 0$. The roots are real and distinct

2. $\\Delta=0$. The roots are real and equal. Their value is $\\displaystyle \\frac{-b}{2a}$

3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.

", "variableReplacementStrategy": "originalfirst", "scripts": {}}], "stepsPenalty": 1}]}, {"name": "Quadratic equations 3", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "emma rand", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/115/"}], "functions": {}, "type": "question", "preamble": {"css": "", "js": ""}, "variablesTest": {"condition": "", "maxRuns": 100}, "tags": ["algebra", "Factorisation", "factorisation", "find roots of a quadratic equation", "quadratic formula", "quadratics", "roots of a quadratic equation", "solving a quadratic equation", "solving equations", "Solving equations", "steps", "Steps"], "variables": {"n4": {"description": "", "definition": "abs(n2)", "group": "Ungrouped variables", "templateType": "anything", "name": "n4"}, "c1": {"description": "", "definition": "switch(f=1, random(1..6),f=2,random(1,3,5,7,9),f=3,random(1,2,5,7,8),f=4,random(1,3,5,7,9),f=6, random(1,5,7,8),f=9,random(1,2,4,7,8),f=8,random(1,3,5,7,9),f=12,random(1,5,7),random(1,3,5,7))", "group": "Ungrouped variables", "templateType": "anything", "name": "c1"}, "rdis": {"description": "", "definition": "switch(disc=0,'The discriminant is '+ 0+' and so we get two repeated roots in this case.',disc<0, 'There are no real roots.','The roots exist and are distinct. ')", "group": "Ungrouped variables", "templateType": "anything", "name": "rdis"}, "rep": {"description": "", "definition": "switch(disc=0,'repeated', ' ')", "group": "Ungrouped variables", "templateType": "anything", "name": "rep"}, "f": {"description": "", "definition": "a*b", "group": "Ungrouped variables", "templateType": "anything", "name": "f"}, "b": {"description": "", "definition": "random(1..4)", "group": "Ungrouped variables", "templateType": "anything", "name": "b"}, "n3": {"description": "", "definition": "2*a*b", "group": "Ungrouped variables", "templateType": "anything", "name": "n3"}, "d": {"description": "", "definition": "if((a*d1)^2=(b*c)^2, max(d1+1,random(1..5))*s3,d1*s3)", "group": "Ungrouped variables", "templateType": "anything", "name": "d"}, "disc": {"description": "", "definition": "(b*c+a*d)^2-4*a*b*c*d", "group": "Ungrouped variables", "templateType": "anything", "name": "disc"}, "n1": {"description": "", "definition": "b*c+a*d", "group": "Ungrouped variables", "templateType": "anything", "name": "n1"}, "c": {"description": "", "definition": "c1*s3", "group": "Ungrouped variables", "templateType": "anything", "name": "c"}, "n2": {"description": "", "definition": "b*c-a*d", "group": "Ungrouped variables", "templateType": "anything", "name": "n2"}, "d1": {"description": "", "definition": "switch(f=1, random(1..6),f=2,random(1,3,5,7,9),f=3,random(1,2,5,7,8),f=4,random(1,3,5,7,9),f=6, random(1,5,7,8),f=9,random(1,2,4,7,8),f=8,random(1,3,5,7,9),f=12,random(1,5,7),random(1,3,5,7))", "group": "Ungrouped variables", "templateType": "anything", "name": "d1"}, "s3": {"description": "", "definition": "random(1,-1)", "group": "Ungrouped variables", "templateType": "anything", "name": "s3"}, "a": {"description": "", "definition": "random(2..5)", "group": "Ungrouped variables", "templateType": "anything", "name": "a"}, "s2": {"description": "", "definition": "random(1,-1)", "group": "Ungrouped variables", "templateType": "anything", "name": "s2"}, "s1": {"description": "", "definition": "random(1,-1)", "group": "Ungrouped variables", "templateType": "anything", "name": "s1"}, "n5": {"description": "", "definition": "a*b", "group": "Ungrouped variables", "templateType": "anything", "name": "n5"}}, "showQuestionGroupNames": false, "variable_groups": [], "statement": "

Find the roots of the following quadratic equation.

", "metadata": {"licence": "None specified", "notes": "", "description": "

Solve for $x$: $\\displaystyle ax ^ 2 + bx + c=0$.

"}, "advice": "

Direct Factorisation

If you can spot a direct factorisation then this is the quickest way to do this question.

For this example we have the factorisation

\\[\\simplify{{a*b} * y ^ 2 + ( {-b*c-a * d}) * y + {c * d} = ({a} * y + { -c}) * ({b} * y + { -d})}\\]

Hence we find the roots:
\\[\\begin{eqnarray} y&=& \\simplify{{n1-n4}/{2*a*b}}\\\\ y&=& \\simplify{{n1+n4}/{2*a*b}} \\end{eqnarray} \\]

Other Methods.

There are several methods of finding the roots – here are the main methods.

Finding the roots of a quadratic using the standard formula.

We can use the following formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

The two roots are

\\[ x = \\frac{-b +\\sqrt{b^2-4ac}}{2a}\\mbox{ and } x = \\frac{-b -\\sqrt{b^2-4ac}}{2a}\\]
there are three main types of solutions depending upon the value of the discriminant $\\Delta=b^2-4ac$

1. $\\Delta \\gt 0$. The roots are real and distinct

2. $\\Delta=0$. The roots are real and equal. Their common value is $\\displaystyle -\\frac{b}{2a}$

3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.

For this question the discriminant of $\\simplify{{a*b}x^2+{-b*c-a*d}x+{c*d}}$ is $\\Delta = \\simplify[std]{{-n1}^2-4*{a*b*c*d}}=\\var{disc}$

{rdis}.

So the {rep} roots are:

Completing the square.

Input numbers as fractions or integers not as a decimals.

Solve for $x$: \\[\\simplify[std]{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d}}=0\\]
The least root is $x=\\;$ [[0]]. The greatest root is $x=\\;$ [[1]]

You can get more information on solving a quadratic by clicking on Show steps. You will lose 1 mark if you do so.

Enter the least root first. If the roots are equal, enter the root in both input boxes.

Enter the roots as fractions or integers, not as decimals.

", "variableReplacementStrategy": "originalfirst", "steps": [{"marks": 0, "type": "information", "variableReplacements": [], "showCorrectAnswer": true, "prompt": "

Finding the roots by factorisation.

Finding a factorisation of a quadratic $q(x)=a(x-r)(x-s)$ where $a$ is the coefficient of $x^2$ gives the roots $x=r$, $x=s$ immediately.

If you cannot find a factorisation then there are several other methods you can use.

Using the formula for the roots.

You can find the roots by using the formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

The two roots are:

\\[ x = \\frac{-b +\\sqrt{b^2-4ac}}{2a}\\mbox{ and } x = \\frac{-b -\\sqrt{b^2-4ac}}{2a}\\]
there are three main types of solutions depending upon the value of the discriminant $\\Delta=b^2-4ac$

1. $\\Delta \\gt 0$. The roots are real and distinct

2. $\\Delta=0$. The roots are real and equal. Their value is $\\displaystyle \\frac{-b}{2a}$

3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.

Find the roots of the following quadratic equation.

", "metadata": {"licence": "None specified", "notes": "", "description": "

Solve for $x$: $\\displaystyle ax ^ 2 + bx + c=0$.

"}, "advice": "

Direct Factorisation

If you can spot a direct factorisation then this is the quickest way to do this question.

For this example we have the factorisation

\\[\\simplify{{a*b} * y ^ 2 + ( {-b*c-a * d}) * y + {c * d} = ({a} * y + { -c}) * ({b} * y + { -d})}\\]

Hence we find the roots:
\\[\\begin{eqnarray} y&=& \\simplify{{n1-n4}/{2*a*b}}\\\\ y&=& \\simplify{{n1+n4}/{2*a*b}} \\end{eqnarray} \\]

Other Methods.

There are several methods of finding the roots – here are the main methods.

Finding the roots of a quadratic using the standard formula.

We can use the following formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

The two roots are

\\[ x = \\frac{-b +\\sqrt{b^2-4ac}}{2a}\\mbox{ and } x = \\frac{-b -\\sqrt{b^2-4ac}}{2a}\\]
there are three main types of solutions depending upon the value of the discriminant $\\Delta=b^2-4ac$

1. $\\Delta \\gt 0$. The roots are real and distinct

2. $\\Delta=0$. The roots are real and equal. Their common value is $\\displaystyle -\\frac{b}{2a}$

3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.

For this question the discriminant of $\\simplify{{a*b}x^2+{-b*c-a*d}x+{c*d}}$ is $\\Delta = \\simplify[std]{{-n1}^2-4*{a*b*c*d}}=\\var{disc}$

{rdis}.

So the {rep} roots are:

Completing the square.

Input numbers as fractions or integers not as a decimals.

Solve for $x$: \\[\\simplify[std]{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d}}=0\\]
The least root is $x=\\;$ [[0]]. The greatest root is $x=\\;$ [[1]]

You can get more information on solving a quadratic by clicking on Show steps. You will lose 1 mark if you do so.

Enter the least root first. If the roots are equal, enter the root in both input boxes.

Enter the roots as fractions or integers, not as decimals.

", "variableReplacementStrategy": "originalfirst", "steps": [{"marks": 0, "type": "information", "variableReplacements": [], "showCorrectAnswer": true, "prompt": "

Finding the roots by factorisation.

Finding a factorisation of a quadratic $q(x)=a(x-r)(x-s)$ where $a$ is the coefficient of $x^2$ gives the roots $x=r$, $x=s$ immediately.

If you cannot find a factorisation then there are several other methods you can use.

Using the formula for the roots.

You can find the roots by using the formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

The two roots are:

\\[ x = \\frac{-b +\\sqrt{b^2-4ac}}{2a}\\mbox{ and } x = \\frac{-b -\\sqrt{b^2-4ac}}{2a}\\]
there are three main types of solutions depending upon the value of the discriminant $\\Delta=b^2-4ac$

1. $\\Delta \\gt 0$. The roots are real and distinct

2. $\\Delta=0$. The roots are real and equal. Their value is $\\displaystyle \\frac{-b}{2a}$

3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.

Find the roots of the following quadratic equation.

", "metadata": {"licence": "None specified", "notes": "", "description": "

Solve for $x$: $\\displaystyle ax ^ 2 + bx + c=0$.

"}, "advice": "

Direct Factorisation

If you can spot a direct factorisation then this is the quickest way to do this question.

For this example we have the factorisation

\\[\\simplify{{a*b} * y ^ 2 + ( {-b*c-a * d}) * y + {c * d} = ({a} * y + { -c}) * ({b} * y + { -d})}\\]

Hence we find the roots:
\\[\\begin{eqnarray} y&=& \\simplify{{n1-n4}/{2*a*b}}\\\\ y&=& \\simplify{{n1+n4}/{2*a*b}} \\end{eqnarray} \\]

Other Methods.

There are several methods of finding the roots – here are the main methods.

Finding the roots of a quadratic using the standard formula.

We can use the following formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

The two roots are

\\[ x = \\frac{-b +\\sqrt{b^2-4ac}}{2a}\\mbox{ and } x = \\frac{-b -\\sqrt{b^2-4ac}}{2a}\\]
there are three main types of solutions depending upon the value of the discriminant $\\Delta=b^2-4ac$

1. $\\Delta \\gt 0$. The roots are real and distinct

2. $\\Delta=0$. The roots are real and equal. Their common value is $\\displaystyle -\\frac{b}{2a}$

3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.

For this question the discriminant of $\\simplify{{a*b}x^2+{-b*c-a*d}x+{c*d}}$ is $\\Delta = \\simplify[std]{{-n1}^2-4*{a*b*c*d}}=\\var{disc}$

{rdis}.

So the {rep} roots are:

Completing the square.

Input numbers as fractions or integers not as a decimals.

Solve for $x$: \\[\\simplify[std]{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d}}=0\\]
The least root is $x=\\;$ [[0]]. The greatest root is $x=\\;$ [[1]]

You can get more information on solving a quadratic by clicking on Show steps. You will lose 1 mark if you do so.

Enter the least root first. If the roots are equal, enter the root in both input boxes.

Enter the roots as fractions or integers, not as decimals.

", "variableReplacementStrategy": "originalfirst", "steps": [{"marks": 0, "type": "information", "variableReplacements": [], "showCorrectAnswer": true, "prompt": "

Finding the roots by factorisation.

Finding a factorisation of a quadratic $q(x)=a(x-r)(x-s)$ where $a$ is the coefficient of $x^2$ gives the roots $x=r$, $x=s$ immediately.

If you cannot find a factorisation then there are several other methods you can use.

Using the formula for the roots.

You can find the roots by using the formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

The two roots are:

\\[ x = \\frac{-b +\\sqrt{b^2-4ac}}{2a}\\mbox{ and } x = \\frac{-b -\\sqrt{b^2-4ac}}{2a}\\]
there are three main types of solutions depending upon the value of the discriminant $\\Delta=b^2-4ac$

1. $\\Delta \\gt 0$. The roots are real and distinct

2. $\\Delta=0$. The roots are real and equal. Their value is $\\displaystyle \\frac{-b}{2a}$

3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.

Find the roots of the following quadratic equation.

", "metadata": {"licence": "None specified", "notes": "", "description": "

Solve for $x$: $\\displaystyle ax ^ 2 + bx + c=0$.

"}, "advice": "

Direct Factorisation

If you can spot a direct factorisation then this is the quickest way to do this question.

For this example we have the factorisation

\\[\\simplify{{a*b} * y ^ 2 + ( {-b*c-a * d}) * y + {c * d} = ({a} * y + { -c}) * ({b} * y + { -d})}\\]

Hence we find the roots:
\\[\\begin{eqnarray} y&=& \\simplify{{n1-n4}/{2*a*b}}\\\\ y&=& \\simplify{{n1+n4}/{2*a*b}} \\end{eqnarray} \\]

Other Methods.

There are several methods of finding the roots – here are the main methods.

Finding the roots of a quadratic using the standard formula.

We can use the following formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

The two roots are

\\[ x = \\frac{-b +\\sqrt{b^2-4ac}}{2a}\\mbox{ and } x = \\frac{-b -\\sqrt{b^2-4ac}}{2a}\\]
there are three main types of solutions depending upon the value of the discriminant $\\Delta=b^2-4ac$

1. $\\Delta \\gt 0$. The roots are real and distinct

2. $\\Delta=0$. The roots are real and equal. Their common value is $\\displaystyle -\\frac{b}{2a}$

3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.

For this question the discriminant of $\\simplify{{a*b}x^2+{-b*c-a*d}x+{c*d}}$ is $\\Delta = \\simplify[std]{{-n1}^2-4*{a*b*c*d}}=\\var{disc}$

{rdis}.

So the {rep} roots are:

Completing the square.

Input numbers as fractions or integers not as a decimals.

Solve for $x$: \\[\\simplify[std]{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d}}=0\\]
The least root is $x=\\;$ [[0]]. The greatest root is $x=\\;$ [[1]]

You can get more information on solving a quadratic by clicking on Show steps. You will lose 1 mark if you do so.

Enter the least root first. If the roots are equal, enter the root in both input boxes.

Enter the roots as fractions or integers, not as decimals.

", "variableReplacementStrategy": "originalfirst", "steps": [{"marks": 0, "type": "information", "variableReplacements": [], "showCorrectAnswer": true, "prompt": "

Finding the roots by factorisation.

Finding a factorisation of a quadratic $q(x)=a(x-r)(x-s)$ where $a$ is the coefficient of $x^2$ gives the roots $x=r$, $x=s$ immediately.

If you cannot find a factorisation then there are several other methods you can use.

Using the formula for the roots.

You can find the roots by using the formula for finding the roots of a general quadratic equation $ax^2+bx+c=0$

The two roots are:

\\[ x = \\frac{-b +\\sqrt{b^2-4ac}}{2a}\\mbox{ and } x = \\frac{-b -\\sqrt{b^2-4ac}}{2a}\\]
there are three main types of solutions depending upon the value of the discriminant $\\Delta=b^2-4ac$

1. $\\Delta \\gt 0$. The roots are real and distinct

2. $\\Delta=0$. The roots are real and equal. Their value is $\\displaystyle \\frac{-b}{2a}$

3. $\\Delta \\lt 0$. There are no real roots. The root are complex and form a complex conjugate pair.

", "variableReplacementStrategy": "originalfirst", "scripts": {}}], "stepsPenalty": 1}]}, {"name": "Quadratic modulus graph", "extensions": ["geogebra"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Adrian Jannetta", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/164/"}], "tags": [], "metadata": {"description": "

This uses an embedded Geogebra graph of a cubic polynomial with random coefficients set by NUMBAS. Student has to decide what kind of map it represents and whether an inverse function exists.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

{geogebra_applet('https://ggbm.at/DN6naqhm', defs, [])}

The graph shows a quadratic function of the form \\[ y = |f(x)| \\] and a straight line $y=\\var{c}$.

", "advice": "

This is not a function. The graph fails the vertical line test, because a vertical line would usually touch two points on the curve.

This is a one-to-many map. That means each input ($x$-axis, domain) often go to two outputs ($y$-axis, range).

This relation does not have an inverse function. Although it passes the horizontal line test....remember, the horizontal line test only applies to functions!

", "rulesets": {}, "variables": {"q2": {"name": "q2", "group": "Ungrouped variables", "definition": "a*b+c", "description": "", "templateType": "anything"}, "x1": {"name": "x1", "group": "Ungrouped variables", "definition": "(-p+sqrt(p^2-4*q1))/2", "description": "", "templateType": "anything"}, "a": {"name": "a", "group": "Ungrouped variables", "definition": "random(-6..6 except 0)", "description": "", "templateType": "anything"}, "p": {"name": "p", "group": "Ungrouped variables", "definition": "-(a+b)", "description": "", "templateType": "anything"}, "b": {"name": "b", "group": "Ungrouped variables", "definition": "random(-6..6 except 0)", "description": "", "templateType": "anything"}, "x3": {"name": "x3", "group": "Ungrouped variables", "definition": "(-p+sqrt(p^2-4*q2))/2", "description": "", "templateType": "anything"}, "x2": {"name": "x2", "group": "Ungrouped variables", "definition": "(-p-sqrt(p^2-4*q1))/2", "description": "", "templateType": "anything"}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "random(1..round(M)-1)", "description": "", "templateType": "anything"}, "M": {"name": "M", "group": "Ungrouped variables", "definition": "abs((a*b)-0.25*(a+b)^2)", "description": "", "templateType": "anything"}, "ans": {"name": "ans", "group": "Ungrouped variables", "definition": "sort([x1,x2,x3,x4])", "description": "", "templateType": "anything"}, "eps": {"name": "eps", "group": "Ungrouped variables", "definition": "0.5*10^(-4)", "description": "", "templateType": "anything"}, "x4": {"name": "x4", "group": "Ungrouped variables", "definition": "(-p-sqrt(p^2-4*q2))/2", "description": "", "templateType": "anything"}, "q1": {"name": "q1", "group": "Ungrouped variables", "definition": "a*b-c", "description": "", "templateType": "anything"}, "defs": {"name": "defs", "group": "Ungrouped variables", "definition": "[\n ['a',a],\n ['b',b],\n ['c',c]\n]", "description": "", "templateType": "anything"}}, "variablesTest": {"condition": "a<>b and M<11 and M>1", "maxRuns": "200"}, "ungrouped_variables": ["a", "b", "defs", "M", "c", "p", "q1", "q2", "x1", "x2", "x3", "x4", "ans", "eps"], "variable_groups": [], "functions": {"graph": {"parameters": [], "type": "html", "language": "jme", "definition": "geogebra_applet('https://ggbm.at/aXxv2ECN', defs, [])"}}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Input the correct expression for $f(x)$ below.

$f(x) = $ [[0]]

", "stepsPenalty": "1", "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Hint: You can see from the graph that $x=\\var{a}$ and $x=\\var{b}$ at $y=0$.

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "alternatives": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "alternativeFeedbackMessage": "", "useAlternativeFeedback": false, "answer": "(x-{a})*({b}-x})", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "valuegenerators": [{"name": "x", "value": ""}]}], "answer": "(x-{a})*(x-{b})", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": true, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "valuegenerators": [{"name": "x", "value": ""}]}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

The line $y=\\var{c}$ intersects the line at four points. Find the $x$-values of those points and list them in ascending numerical order (smallest value first).

$x = $ [[0]]

$x = $ [[1]]

$x = $ [[2]]

$x = $ [[3]]

(Give answers to 4 decimal places).

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