Question
Solve the following quadratic equation by factorization.
$x^2+x-20=0$
$x^2+x-20=0$
✓
Answer
$x^2+x-20=0 $
$\Rightarrow x^2+5 x-4 x-20=0$
$\Rightarrow x(x+5)-4(x+5)=0$
$\Rightarrow(x+5)(x-4)=0 $
$x+5=0 \Rightarrow x=-5$
$x-4=0 \Rightarrow x=4$
Hence, $x =-5$ and $x =4$ are roots of the equation.
$\Rightarrow x^2+5 x-4 x-20=0$
$\Rightarrow x(x+5)-4(x+5)=0$
$\Rightarrow(x+5)(x-4)=0 $
$x+5=0 \Rightarrow x=-5$
$x-4=0 \Rightarrow x=4$
Hence, $x =-5$ and $x =4$ are roots of the equation.
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