// Numbas version: finer_feedback_settings {"name": "Practice Test FY001", "metadata": {"description": "

NUMBAS practice exam for FY001 - Your score won't be counted for your mark.

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Repeated roots questions, using the equation from Part a:

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$\\simplify{{f1}x^2+{f2}*x+{f3}}=0$

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These are solved in the exact same way as the previous set of quadratics, but the 'roots' (solutions) are the same.

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This type of equation is described as having 'repeated roots'.

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The graphed functions would only touch the $x$-axis once either at their peak or trough.

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Looking at the above equation, you may be able to instantly see that the brackets must both include the quantity $\\var{c}$.

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If not, don't worry, just go through the same process of trial and error as taught previously until you reach your solution of

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$\\simplify{({a}x+{c})({b}x+{d})}$.

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Then, noticing this can be further simplified, write the final simplification as:

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$\\simplify{({a}x+{c})^2}$.

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\n

\n

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Note: If the $x^2$ term has a coefficient and simplifies to, for example, $(2x+4)(x+4)$, this is not defined as having repeated roots, since the solutions are:

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$2x+4=0$  so   $x=\\frac{-4}{2}=-2$

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and

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$x+4=0$    so   $x=-4$

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so are in fact different.

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\n

\n

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Zero constants equations, using Part d:

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$\\simplify{{f14}x^2+{f24}*x+{f34}}=0$

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Some equations will not have a constant at the end, meaning one of the constants is zero.

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In this case, if there is still an $x$ term, you know one of the constants must still be non-zero.

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You must therefore, take out a factor of a number $\\times x$.

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Here, the factor could be $\\simplify{{a4}x}$ or $\\simplify{{b4}x}$.

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It doesn't matter which you choose, but in this example we will use $\\simplify{{a4}x}$.

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In order to pull out this factor, the rest of the equation must be divided by $\\simplify{{a4}x}$:

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$\\frac{\\simplify{{f14}x^2+{f24}*x+{f34}}}{\\simplify{{a4}*x}}=\\simplify{{f14}x/{a4}+{f24}/{a4}}$.

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Therefore, the solution is:

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$\\simplify{{a4}x*({f14}x/{a4}+{f24}/{a4})}$.

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$\\simplify{{f1}x^2+{f2}*x+{f3}}$

\n

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The answer must be factorised.

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The answer must be factorised.

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$\\simplify{{f12}x^2+{f22}*x+{f32}}$

\n

$=$[[0]] 

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The answer must be factorised.

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$\\simplify{{f13}x^2+{f23}*x+{f33}}$

\n

$=$[[0]] 

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The answer must be factorised.

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$\\simplify{{f14}x^2+{f24}*x+{f34}}$

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$=$[[0]] 

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The answer must be factorised.

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Hint: Remember one or more of the constants could be zero.

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Simplify the following quadratics into two linear factors in the form $(ax+c)(bx+d)$.

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Quadratics are defined as having 'repeated roots' if the both roots in the solution are the same. This can be written as $(x+a)(x+a)$, but can be simplified to $(x+a)^2$.

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Factorising further basic quadratics into linear expressions

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Practice with adding, subtracting and dividing basic algebraic fractions

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Express the following as a single fraction and simplify where possible.

\n

Remember to use brackets appropriately and to use a * for multiplication for letters.

\n

-----------------------------------

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See Lecture 4.1 and Workshop 4.2 for background and examples.

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For background on fractions see 2.3.  For background on algebraic manipulations see 2.6 and 3.3.

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a/x  + 1/bx

\n

c/x - a/d

\n

a/x * d/by

\n

d/x  /   c/ay

\n

a/x -  1 / (c-fx)

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c/(d + y) - (b)/(a+ x)

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\\[ \\frac{\\var{a}}{x} + \\frac{1}{\\var{b}x} \\]

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\\[ \\frac{\\var{c}}{x} - \\frac{\\var{a}}{\\var{d}} \\]

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\\[ \\frac{\\var{a}}{x} \\times \\frac{\\var{d}}{\\var{b}y} \\]

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\\[ \\frac{\\var{d}}{x} \\div \\frac{\\var{c}}{\\var{a}y} \\]

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\\[ \\dfrac{\\var{a}}{x} + \\dfrac{1}{(\\var{c} -\\var{f}x)} \\]

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\\[ \\frac{\\var{c}}{(\\var{d} + y)} - \\frac{\\var{b}}{(\\var{a} + x)}\\]

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Have you checked that you have entered answers appropriately: used brackets and * where necessary.

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Simplifying indices.

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Use the following indices law to help answer this question:

\n

$x^a \\times x^b = x^{a+b}$

\n

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$x^\\var{a} \\times x^\\var{b}$

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Use the following law to help answer this question:

\n

$x^a \\times x^b = x^{a+b}$

\n

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$p^\\var{c} \\times p^\\var{d}$

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Use the following law to answer this question:

\n

$(ax^b)^c = a^cx^{bc}$

\n

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$(\\var{a}k^\\var{b})^\\var{f}$

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Use the following law:

\n

$x^a \\times x^b = x^{a+b}$

\n

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$y^{1/\\var{a}} \\times y^{1/\\var{b}}$

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Use the following law:

\n

$x^a \\div x^b = x^{a-b}$

\n

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$c^\\var{a}$$c^\\var{b}$

\n

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Use the following law to answer this question:

\n

$\\frac{ax^c}{bx^d}= \\frac{a}{b}x^{(c-d)}$

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$\\var{a}h^\\var{c}$$\\var{b}h^\\var{d}$

\n

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This question differs from part f due to the brackets. Using principles of BODMAS, the brackets need to be expanded first. 

\n

$(4d)^\\var{g}$ expands to $4^\\var{g}d^\\var{g}$ and $(2d)^\\var{h}$ expands to $2^\\var{h}d^\\var{h}$

\n

Now you are left with a simple division question as follows:

\n

$\\frac{4^{\\var{g}}d^{\\var{g}}}{2^{\\var{h}}d^{\\var{h}}}$

\n

\n

Use the principle:

\n

$\\frac{ax^c}{bx^d}= \\frac{a}{b}x^{(c-d)}$ to answer the question.

\n

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$(4d)^\\var{g}$$(2d)^\\var{h}$

\n

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Simplify each of the following expressions, giving your answer in its simplest form.

\n

Click 'Show steps' for guidance on which index law is applicable.

", "ungrouped_variables": ["a", "b", "c", "d", "f", "g", "h", "j"], "advice": "

Recall the laws of indices to help solve the problems:

\n

$x^a \\times x^b = x^{a+b}$

\n

$x^a \\div x^b = x^{a-b}$

\n

$x^{-a} = \\frac{1}{x^a}$

\n

$(x^a)^b = x^{ab}$

\n

$(\\frac{x}{y})^a = \\frac{x^a}{y^a}$

\n

$x^\\frac{a}{b} = (\\sqrt[b]{x})^{a}$

\n

$x^0 = 1$

\n

\n

Worked Solutions:

\n

Part a)               $x^{(\\var{a}+\\var{b})}=\\simplify{x^{({a}+{b})}}$

\n

Part b)               $p^{(\\var{c}+\\var{d})}=\\simplify{p^{({c}+{d})}}$

\n

Part c)               $\\var{a}^\\var{f}\\times{k^{(\\var{b}\\times\\var{f})}}=\\simplify{{a}^{f}*k^{({b}*{f})}}$

\n

Part d)               $y^{((\\var{a}+\\var{b})/(\\var{a}\\times\\var{b}))}=y^{\\frac{\\simplify{{a}+{b}}}{\\simplify{{a}*{b}}}}$

\n

Part e)               $c^{(\\var{a}-\\var{b})}=c^\\simplify{({a}-{b})}$

\n

Part f)                $\\frac{\\var{a}}{\\var{b}}h^{\\var{c}-\\var{d}}=\\frac{\\var{a}}{\\var{b}}{\\simplify{h^{{c}-{d}}}}$

\n

Part g)               $\\frac{4^\\var{g}}{2^\\var{h}}\\times{d^{\\var{g}-\\var{h}}}=\\simplify{(4^{g})/(2^{h})*d^{g-h}}$

\n

Part h)               $\\frac{6^\\var{g}}{9^\\var{h}}\\times{p^{\\var{h}\\var{j}-\\var{g}\\var{f}}}=\\simplify{(6^{g})/(9^{h})*p^{h*j-g*f}}$

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Finding the roots by factorisation.

\n

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.

\n

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

\n

Using the formula for the roots.

\n

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

\n

The two roots are:

\n

\\[ 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 $\\Delta=b^2-4ac$

\n

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

\n

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

\n

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

\n

 

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Solve for $x$:

\n

\\[\\simplify[std]{{a*b} * x ^ 2 + ( {-b*c-a * d}) * x + {c * d}}=0\\]

\n

$x=$ [[0]] or [[1]].

\n

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

\n

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

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Input numbers as fractions or integers not as a decimals.

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Input numbers as fractions or integers not as a decimals.

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Find the roots of the following quadratic equation.

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

\n

Entering the correct roots in any order is marked as correct. However, entering one correct and the other incorrect gives feedback stating that both are incorrect.

"}, "advice": "\n\t

Direct Factorisation

\n\t

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

\n\t

For this example we have the factorisation

\n\t

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

\n\t

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

\n\t

Other Methods.

\n\t

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

\n\t

Finding the roots of a quadratic using the standard formula.

\n\t

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

\n\t

The two roots are

\n\t

\\[ 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 $\\Delta=b^2-4ac$

\n\t

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

\n\t

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

\n\t

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

\n\t

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

\n\t

{rdis}.

\n\t

So the {rep} roots are:

\n\t

\\[\\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}\\]

\n\t

Completing the square.

\n\t

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}\\]

\n\t \n\t \n\t \n\t \n\t \n\t \n\t \n\t \n\t \n\t \n\t \n\t"}]}], "allowPrinting": true, "navigation": {"allowregen": false, "reverse": true, "browse": true, "allowsteps": false, "showfrontpage": true, "showresultspage": "review", "navigatemode": "sequence", "onleave": {"action": "none", "message": ""}, "preventleave": true, "startpassword": ""}, "timing": {"allowPause": false, "timeout": {"action": "warn", "message": "

YOU HAVE RUN OUT OF TIME

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LAST 5 MINUTES

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This is Test for you to practice your knowledge of the material, and also usage of NUMBAS system in exam conditions. It won't be counted towards your final grade. 

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