Question

Find the root of the following quadratic equations if they exist by the method of completing square.

(i) 2x^2 + x + 4 = 0​

Answer 1

$\huge\underline\mathbb{SOLUTION:-}$

$\mathsf \red {(i)\:2x^2 + x + 4 = 0}$

$\mathsf {Dividing\:equation\:by\:2.}$

$\mathsf {x^2 + \frac{x}{2} + 2 = 0}$

$\mathsf {Following\:the\:procedure\:of\:completing\:square,}$

$\implies \mathsf {x^2 + \frac{x}{2} + 2 + \bigg(\frac{1}{2}\bigg)^2 - \bigg(\frac{1}{4}\bigg)^2 = 0}$

$\implies \mathsf {x^2 + \bigg(\frac{1}{4}\bigg)^2 + \frac{x}{2} + 2 - \bigg(\frac{1}{4}\bigg)^2 = 0}$

• [(a + b)² = a² + b² + 2ab]

$\implies \mathsf {\bigg(x + \frac{1}{4}\bigg)^2 + 2 - \frac{1}{16} = 0}$

$\implies \mathsf {\bigg(x + \frac{1}{4}\bigg)^2 = \frac{1}{16} - 2 = \frac{1 - 32}{16} }$

$\mathsf {Taking\:square\:root\:on\:both\:sides.}$

• Right hand side does not exist because square root of negative number does not exist.

$\therefore \mathsf \blue {There\:is\:no\:solution\:for\:quadratic\:equation\:2x^2 + x + 4 = 0}$

Answer 2

Given Quadratic equation ;

2x^2 + x +4 = 0

•2x^2 +x = -4

divide 2 on both sides, we get

2x^2/2 + x/2 = -4/2

so ,we get

x^2 + x/2 = -2

Add [1/4]^2 on both sides, we get

x^2 + x/2 +[1/4]^2 =-2 +[1/4]^2

[x+1/4]^2 = -2 +1/16

=-32+1/16

= -33/16

x+1/4 =√-33/16 =√-33/4

x= -1/4 - √-33/4,-1/4+√-33/4

● X = 1+√33/4,1-√33/4

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