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a quick test to see what the students see. 

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

$f(x) = \\simplify{{c1}x^2 + {c2}x + {c3}}$

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50% chance of being the same as root1

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How many distinct solutions does $f(x)=0$ have?

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Write the two roots, $x_1$ and $x_2$, of the equation $f(x)=0$, such that $x_1 \\lt x_2$:

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$x_1 = $ [[0]]

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$x_2 = $ [[1]]

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What is the discriminant of $f(x)$?

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For a quadratic equation in the form $ax^2 + bx + c$, the discriminant $D$ is

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\\[ D = b^2-4ac \\]

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The number of distinct roots of the equation depends on the value of the discriminant:

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The number of apartments in a housing development has been increasing by a constant amount every year. At the end of the first year, the number of apartments was {u_1}, and at the end of the {years} year, the number of apartments was {u_years}. The number of apartments, $y$, can be determined by the equation $y = mt + n$,  where $t$ is the time, in years.

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$m$ is the gradient of the line, it is also the common difference in the arithmetic sequence so you can use the formula $u_n = u_1 + (n-1)d$

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In this case $n =$  {n} so {u_years} = {u_1} + ({n}-1) $\\times d$ 

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m = {u_years - u_1} $\\div$ {n-1}

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m = {d}

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$n$ represents the situation at the beginning of year 1, so is $u_1 - d$

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$n =$ {u_1} - {d}

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$n =$ {u_0}

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first term (houses at end of first year)

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conversion list for number of years to english

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number of years at which second number of houses is given

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n converted to word

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u_n number of houses after n years

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common difference to apply

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first term of sequence (houses at start of first year)

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Find the value of $m$.

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Find the value of $n$.

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A new way for solving Quadratic Equation:

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The quadratic equation $ax^2+bx+c=0$ can be solved in a different and easy way as follows:

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Let $\\alpha$ and $\\beta$ be the two roots.  Then we know $\\alpha\\times\\beta=\\dfrac{c}{a}$ and $\\alpha+\\beta=-\\dfrac{b}{a}$

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We shall use the property that the roots are equidistant from their average.

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So we can take them as $\\alpha=\\dfrac{1}{2}\\rm{sum}-u$ and $\\beta=\\dfrac{1}{2}\\rm{sum}+u$ Or $\\boxed{\\alpha=-\\dfrac{1}{2}\\dfrac{b}{a}-u}$ and $\\boxed{\\beta=-\\dfrac{1}{2}\\dfrac{b}{a}+u}$ .

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Consequently we have $\\alpha\\times\\beta=\\left(-\\dfrac{1}{2}\\dfrac{b}{a}-u\\right)\\times\\left(-\\dfrac{1}{2}\\dfrac{b}{a}+u\\right)$

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$\\implies \\dfrac{c}{a}=\\dfrac{1}{4}\\dfrac{b^2}{a^2}-u^2\\implies u^2=\\dfrac{1}{4}\\dfrac{b^2}{a^2}-\\dfrac{c}{a}$.

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Simplifying $u^2=\\dfrac{b^2-4ac}{4a^2}\\implies \\boxed{u=\\dfrac{\\sqrt{b^2-4ac}}{2a}}$ .

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Substituting above we get

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$\\alpha=\\dfrac{-b-\\sqrt{b^2-4ac}}{2a}$ and $\\beta=\\dfrac{-b+\\sqrt{b^2-4ac}}{2a}$.

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Note : $\\boxed{\\alpha=\\rm{average}-u}$ and $\\boxed{\\beta=\\rm{average}+u}$

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______________________________________________________________________________________________________________________________

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Let us consider the following example:

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$x^2-5x+6=0$ .

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We generally start from the product $6$ and find factors by inspection so that the sum will be $5$. In stead, in this method we start from the sum and find $u$.

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Now sum of roots$=5\\implies \\rm{average}=\\dfrac{5}{2}$.

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So roots are $\\dfrac{5}{2}-u$ and $\\dfrac{5}{2}+u$ .

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Therefore product of roots $=\\dfrac{25}{4}-u^2\\implies 6=\\dfrac{25}{4}-u^2\\implies u^2=\\dfrac{25}{4}-6\\implies u^2=\\dfrac{1}{4}\\implies u=\\pm\\dfrac{1}{2}$.

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So roots are $\\dfrac{5}{2}-\\dfrac{1}{2}$ and $\\dfrac{5}{2}+\\dfrac{1}{2}$ . Or roots are $2$ and $3$. 

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______________________________________________________________________________________________________________________________               

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First coefficient

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random

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Sum of the roots

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product of the roots

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Distance of the root from the average

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Average of the roots

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Average-u

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Average+u

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Discriminant of the equation

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Increment on a

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PRACTICE EXERCISE:

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Solve the equation $\\simplify{{a}x^2+{b}x+{c}=0}$

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Fill in the gaps below.

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a=[[0]]

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b=[[1]]

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c=[[2]]

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Sum of roots=[[3]]

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Avg(Average of roots)=[[4]]

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$b^2$=[[5]]

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$4ac$=[[6]]

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Discriminant $(b^2-4ac)$=[[7]]

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$u\\left(=\\dfrac{\\sqrt{b^2-4ac}}{2a}\\right)$=[[8]]

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$\\alpha(=Avg-u)$=[[9]]

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$\\beta(=Avg+u)$=[[10]]

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{"description": "

All the answers in this question are equations. In order to mark each\n equation, Numbas needs to pick some values that satisfy the equation \nand some that don't, and check that the student's answer agrees with the\n expected answer.

Any equation with the same solution set as the expected answer will be marked correct.

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

All the answers in this question are equations. In order to mark each equation, Numbas needs to pick some values that satisfy the equation and some that don't, and check that the student's answer agrees with the expected answer.

\n

Any equation with the same solution set as the expected answer will be marked correct.

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$s = \\simplify[]{{factor}p}$

\n

Also try $s-\\var{factor}p=0$ and $p=s/\\var{factor}$.

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$s = \\simplify[]{{factor}p}$

\n

But this time there's a pattern-match restriction that your answer must be of the form $s = \\ldots$.

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$x^2+y^2=1$

\n

Also try $1-x^2-y^2=0$ and $y=\\sqrt{1-x^2}$.

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$x^2 + y^2 + z^2=1$

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$x^2+y^2>1$

\n

Because this is an inequality, we need to randomly pick values either side of the critical curve, and some exactly on it.

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