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Question

Solve the following quadratic equations by factorisation:
$$\left( i \right)\left( {x + 1} \right)\left( {2x + 1} \right) = 15$$
$$\left( {ii} \right)2{x^2} + 5 = 17x$$
$$\left( {iii} \right)x\left( {x + 2} \right) + \left( {x + 1} \right)\left( {2x - 1} \right) = 17$$
$$\left( {iv} \right){x^2} - x - 20 = 0$$


Solution

$$(1) (x+1)(2x+1)=15$$
$$\Rightarrow 2x^{2}+2x+x+1=15$$
$$\Rightarrow 2x^{2}+3\times -14=0$$
$$\Rightarrow 2x^{2}+7x-4x-14=0$$
$$\Rightarrow x(2x+7)-2(2x+7)=0$$
$$\Rightarrow (x-2)(2x+7)=0$$
$$x=2,\frac{-7}{2}$$

$$(2) 12x^{2}-17x+5=0$$
$$12x^{2}-12x-5x+5=0$$
$$12x(x-1)-5(x-1)=0$$
$$(12x-5)(x-1)=0$$
$$x=\frac{5}{12},1$$

$$(3)x^{2}+2x+2x^{2}-x+2x-1=17$$
$$3x^{2}+3x-18=0$$
$$x^{2}+x-6=0$$
$$x^{2}+3x-2x-6=0$$
$$x(x+3)-2(x+3)=0$$
$$(x+3)(x-2)=0$$

$$(4) x^{2}-x-20=0$$
$$x^{2}-4x+5x-20=0$$
$$x(x-4)+5(x-4)=0$$
$$(x+5)(x-4)=0$$

1214625_1428463_ans_edb00b89f961477c9479df7b240489bb.jpg

Mathematics

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