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When we expand a factorised quadratic expression we obtain
\n\\[(x+a)(x+b)=x^2+(a+b)x+ab\\text{.}\\]
\nThis means when factorising the quadratic expression
\n\\[\\simplify{x^2+{r1+r2}x+{r1*r2}=0}\\]
\nwe need to find two values that add together to make $\\var{r1+r2}$ and multiply together to make $\\var{r1*r2}$.
\n\\[\\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
\n\\[\\simplify{(x+{r1})(x+{r2})}=0\\text{.}\\]
\n\nIn 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 $\\var{r2}$.
\nThis would give us a factorised equation of
\n\\[\\simplify {(x+{r1})(x+{r2})}=0\\text{.}\\]
\nIn order to solve for $x$ we need $x$ values that would mean that
\n\\[\\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 $0\\times(x+a)=0$.
\nTherefore our possible $x$ values are
\n\\[\\begin{align}
x_1&=\\var{-r2}\\\\
x_2&=\\var{-r1}\\text{.}
\\end{align}\\]
Andregradslikningen $ax^2+bx+c=0$ har løsningene
\n\\[x=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a},\\space \\text{når}\\space b^2-4ac\\geq0\\text{.}\\]
\nDersom andregradsuttrykket $ax^2+bx+c$ har nullpunktene $x=x_{1}$ og $x=x_{2}$ kan det faktoriseres til $a(x-x_{1})(x-x_{2}).$
Løs andregradslikningen $\\simplify{x^2 +{r1+r2}x+{r1*r2}}=0\\text{.}$
\nInput your values as $x_1$ and $x_2$, where $x_1<x_2$. skriv detta tydligt!
\n$x_1=$ [[0]]
\n$x_2=$ [[1]]
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\n[[0]] $=0$
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