Question

Solve the following quadratic equations by factorisation:
$\left(i\right)\left(x+1\right)\left(2x+1\right)=15$

$\left(ii\right)2x^2+5=17x$

$\left(iii\right)x\left(x+2\right)+\left(x+1\right)\left(2x-1\right)=17$

$\left(iv\right)x^2-x-20=0$

Answer 1

$\left(1\right)(x+1)(2x+1)=15$

$\Rightarrow 2x^2+2x+x+1=15$

$\Rightarrow 2x^2+3\times -14=0$

$\Rightarrow 2x^2+7x-4x-14=0$

$\Rightarrow x(2x+7)-2(2x+7)=0$

$\Rightarrow (x-2)(2x+7)=0$

$x=2,\frac{-7}{2}$

$\left(2\right)12x^2-17x+5=0$

$12x^2-12x-5x+5=0$

$12x(x-1)-5(x-1)=0$

$(12x-5)(x-1)=0$

$x=\frac{5}{12},1$

$\left(3\right)x^2+2x+2x^2-x+2x-1=17$

$3x^2+3x-18=0$

$x^2+x-6=0$

$x^2+3x-2x-6=0$

$x(x+3)-2(x+3)=0$

$(x+3)(x-2)=0$

$\left(4\right)x^2-x-20=0$

$x^2-4x+5x-20=0$

$x(x-4)+5(x-4)=0$

$(x+5)(x-4)=0$