Question

$c) \: solve \: the \: quadratic \: equations.$
$1) \: {x}^{2} - 5x = 0$
$2) {x}^{2} - 12x = 0$
$3) \: {2y}^{2} - 4y = 0$
$4) \: {3x}^{2} + 3x = 0$
$5) {9x}^{2} - 144 = 0$
$6) {x}^{2} - 21 = 0$
$7) \: {4x}^{2} - 100 = 0$
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Answer 1

$\huge\boxed{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{W}\blue{E}\pink{R}}}}$

${\bf{\green{1)x^2-5x=0}}}$

${\bf{\green{=>x(x-5)=0}}}$

${\bf{\green{=>x=0(or)x=5}}}$

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${\bf{\red{2)x^2-12x=0}}}$

${\bf{\red{=>x(x-12)=0}}}$

${\bf{\red{=>x=0 (or) x=12}}}$

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${\bf{\blue{3)2y^2-4y=0}}}$

${\bf{\blue{=>2y(y-2)=0}}}$

${\bf{\blue{=>2y=0(or)y=2}}}$

${\bf{\blue{=>y=0 (or) y=2}}}$

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${\bf{\pink{4)3x^3+3x=0}}}$

${\bf{\pink{=>3x(x+1)=0}}}$

${\bf{\pink{=>3x=0(or)x+1=0}}}$

${\bf{\pink{=>x=0(or)x=-1}}}$

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${\bf{\green{5)9x^2-144=0}}}$

${\bf{\green{=>x^2=144/9}}}$

${\bf{\green{=>x=+12/3(or)x=-12/3}}}$

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${\bf{\orange{6)x^2-21=0}}}$

${\bf{\orange{=>x^2=21}}}$

${\bf{\orange{=>x=+sqrt21(or) x=-sqrt21}}}$

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${\bf{\blue{7)4x^2-100=0}}}$

${\bf{\blue{=>x^2=100/4}}}$

${\bf{\blue{=>x=+10/2(or)x=>-10/2}}}$

Answer 2

Answer:

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