Question

Find the roots of the quadratic equation 2x²+x-4=0 by the method of completing the square method

Answer 1

$\large{\underline{\bf{\blue{Given:-}}}}$

• ✦ 2x² + x -4 = 0

$\large{\underline{\bf{\blue{To\:Find:-}}}}$

✦ we need to find the roots of the equation by completing the square method.

$\huge{\underline{\bf{\green{Solution:-}}}}$

2x² + x -4 = 0

Divide both side by 2.

⠀⠀$\mapsto \rm x {}^{2} + \frac{x}{2} -2 = 0 \\ \\ \rm \: add \: \: ( \frac{1}{4} ) {}^{2} \: on \: both \: sides \\ \\ \rm \: \mapsto \: {x}^{2} + \frac{x}{2} + \frac{1}{16} = 2 + \frac{1}{16} \\ \\ \rm \: \mapsto \:( x + \frac{1}{4} ) {}^{2} = \frac{32 + 1}{16} \\ \\ \rm \: \mapsto(x + \frac{1}{4}) {}^{2} = \frac{33}{16} \\ \\\rm \: \mapsto \: x + \frac{1}{4} = \sqrt{ \frac{33}{16} } \\ \\ \rm \: \mapsto \: x = \frac{ \pm \sqrt{ 33}}{4} - \frac{1}{4} \\ \\\bf \: \mapsto \pink{x = \frac{ \sqrt{33} - 1}{4} } \: \: or \: \: \pink{x = \frac{ - ( \sqrt{33} + 1 )}{4} }$

⠀⠀⠀

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Answer 2

Answer: divide the whole equation by 2

(2x^2 +x-4) /2 =0

x^2 +x/2 -2 =0     .....     (i)

we know  (a+b)^2 =a^2+2ab +b^2

here a=x and 2ab=x/2

          !

=>    2xb=x/2

2b=1/2

b=1/2 x 1/2 =1/4

in (i) adding and subtracting (1/4)^2

x^2 +x/2 -2 +(1/4)^2-(1/4)^2=0

x^2 + x/2 + (1/4)^2 - 2 - (1/4)^2= 0

(x+1/4)^2 - 2 - (1/4)^2 = 0

(x+1/4)^2 -2 -1/16 = 0

(x +1/4)^2 = 2 + 1/16

(x +1/4)^2 =(2(16) +1 )/16

(x+1/4)^2 = (32+1)/16

=> 33/16

=>x+1/4 = + root 33 / 4  

x = root 33/4-1/4

x=(root 33 -1)/4

and x +1/4 = - root 33/4

x= -rt33/4 -1/4

x= -rt33- 1/4

x = -(rt33 +1)/4

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