Question

Find the roots of the Following equation:

(i) $x^2-3x-4=0$ By Factorization

(ii) $5x^2-6x=8$ By Completing squares

(iii) $4x^2-2x=5$ By Completing squares

Answer 1

(i) $x^2-3x-4=0(x+1)(x-4)=0x+1=0orx-4=0x=-1orx=4$

The solution is x = –1, 4

(ii) $5x^2-6x=8$

$x^2-\frac{6}{5}x=\frac{8}{5}$

$x^2-\frac{6}{5}x+\frac{9}{25}=\frac{8}{5}+\frac{9}{25}$

$x^2–\frac{6}{5}x+\frac{9}{25}=\frac{49}{25}$

$(x-\frac{3}{5}{)}^2=\frac{49}{25}$

$x-\frac{3}{5}=\pm \frac{7}{5}$

$x=2andx=-\frac{4}{5}$

(iii) $4x^2-2x=5$

$x^2-\frac{1}{2}x=\frac{5}{4}$

$x^2-\frac{1}{2}x+\frac{1}{16}=\frac{5}{4}+\frac{1}{16}$

${(x-\frac{1}{4})}^2=\frac{21}{16}$

$x=\frac{(1-\sqrt{21})}{4}andx=\frac{(1+\sqrt{21})}{4}$