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This question tests the students ability to factorise simple quadratic equations (where the coefficient of the x^2 term is 1) and use the factorised equation to solve the equation when it is equal to 0.

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a)

When we expand a factorised quadratic expression we obtain

\\[(x+a)(x+b)=x^2+(a+b)x+ab\\text{.}\\]

This means when factorising the quadratic expression

\\[\\simplify{x^2+{r1+r2}x+{r1*r2}=0}\\]

we need to find two values that add together to make $\\var{r1+r2}$ and multiply together to make $\\var{r1*r2}$.

\\[\\begin{align}
\\var{r1} \\times \\var{r2}&=\\var{r1*r2}\\\\
\\var{r1}+\\var{r2}&=\\var{r1+r2}\\\\
\\end{align} \\]

Therefore using $\\var{r1}$ and $\\var{r2}$ we can write out the factorised equation

\\[\\simplify{(x+{r1})(x+{r2})}=0\\text{.}\\]

b)

In order to find the values of $x$ we need to factorise the questions like in part a). To do this we need to find factors of $\\var{r1*r2}$ that add together to give $\\var{r1+r2}$, which could be $\\var{r1}$ and $\\var{r2}$.

This would give us a factorised equation of

\\[\\simplify {(x+{r1})(x+{r2})}=0\\text{.}\\]

In order to solve for $x$ we need $x$ values that would mean that

\\[\\begin{align}
(\\simplify {(x+{r1})})&=0 \\\\
\\text{or}&\\\\
(\\simplify{(x+{r2})})&=0\\text{.}
\\end{align}\\]

If at least one of the factorised brackets equals $0$ then our equation is satisfied because $0\\times(x+a)=0$.

Therefore our possible $x$ values are

\\[\\begin{align}
x_1&=\\var{-r2}\\\\
x_2&=\\var{-r1}\\text{.}
\\end{align}\\]

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Quadratic equations can be written in the form
\\[ax^2+bx+c=0\\text{.}\\]
When $a=1$, these equations can be solved by factorising to create an equation of the form
\\[(x+m)(x+n)=0\\text{,}\\]
where the solutions for $x$ are $-m$ and $-n$.

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Factorise the following equation,

\\[\\simplify{x^2 +{r1+r2}x+{r1*r2}}=0\\text{.}\\]

[[0]] $=0$

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Use your factorised equation to find two values of $x$ that satisfy the equation

\\[\\simplify{x^2 +{r1+r2}x+{r1*r2}}=0\\text{.}\\]

Input your values as $x_1$ and $x_2$, where $x_1<x_2$.

$x_1=$ [[0]]

$x_2=$ [[1]]

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