Question

Solve the following quadratic equations by factorisation :
$\left(i\right)x^2-5x=0\left(ii\right)x^2+6x=27\left(iii\right)9x^2+16=-24x\left(iv\right)121=x^2$

Answer 1

(1) $x^2-5x=0$

$x(x-5)=0\Rightarrow x=0,5$

$\left(2\right)x^2+6x-27=0$

$x^2+9x-3x-27=0$

$x(x+9)-3(x+9)=0$

$(x-3)(x+9)=0\Rightarrow x=3,-9$

$\left(3\right)9x^2+24x+16=0$

$(3x+4{)}^2=0$

(4) $121=x^2$

$x=\pm 11$