Question

Find the roots of the quadratic equation 2x²+x-4=0 by the method equation 2x4+x-4=0 by the method of completing the square?​

Answer 1

Solution

$2 {x}^{2} + x - 4 = 0$

$\implies 2 {x}^{2} + x = 4$

Dividing throughout by '2'

$\implies \dfrac{2 {x}^{2} }{2} + \dfrac{x}{2} = \dfrac{4}{2}$

$\implies {x}^{2} + \dfrac{x}{2} = 2$

It can be written as

$\implies (x)^{2} + 2(x) \bigg( \dfrac{1}{4} \bigg) = 2$

Adding (1/4)² on both sides

$\implies (x)^{2} + 2(x) \bigg( \dfrac{1}{4} \bigg) + \bigg( \dfrac{1}{4} \bigg)^{2} = 2 + \bigg( \dfrac{1}{4} \bigg)^{2}$

$\implies \bigg( x + \dfrac{1}{4} \bigg)^{2} = 2 + \dfrac{1}{ {4}^{2} }$

[ Because a² + b² + 2ab = (a + b)² ]

$\implies \bigg( x + \dfrac{1}{4} \bigg)^{2} = 2 + \dfrac{1}{ 16 }$

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

$\implies \bigg( x + \dfrac{1}{4} \bigg)^{2} = \dfrac{33}{ 16 }$

Taking square root on both sides

$\implies \sqrt{ \bigg( x + \dfrac{1}{4} \bigg)^{2}} = \pm \sqrt{\dfrac{33}{ 16 } }$

$\implies x + \dfrac{1}{4} = \pm \dfrac{ \sqrt{33} }{ \sqrt{16} }$

$\implies x + \dfrac{1}{4} = \pm \dfrac{ \sqrt{33} }{ 4 }$

$\implies x + \dfrac{1}{4} = \dfrac{ \sqrt{33} }{ 4 } \quad or \quad x + \dfrac{1}{4} = - \dfrac{ \sqrt{33} }{4}$

$\implies x = \dfrac{ \sqrt{33} }{ 4 } - \dfrac{1}{4} \quad or \quad x = - \dfrac{ \sqrt{33} }{4} - \dfrac{1}{4}$

$\implies x = \dfrac{ \sqrt{33} - 1 }{ 4 } \quad or \quad x = - \dfrac{ \sqrt{33} + 1 }{4}$

Hence, (√33 - 1)/4 and - (√33 + 1)/4 are the roots of the equation.

Answer 2

Answer:

$\large\bold\red{Roots = \frac{ - 1 + \sqrt{33} }{4} , \frac{ - 1 -\sqrt{33} }{4} }$

Step-by-step explanation:

Given,

A quadratic equation ,

$2 {x}^{2} + x - 4 = 0$

But,

we know that,

The standard form of a quadratic equations is,

$\large \boxed{a {x}^{2} + bx + c = 0}$

Comparing the Coefficients,

We get,

$a = 2 \\ b = 1 \\ c = - 1$

Now,

We have to find it's roots by completing the square method,

For that,

Divide the whole Equation by 2,

Therefore,

We get,

$= > \frac{2 {x}^{2} + x - 4 }{2} = \frac{0}{2} \\ \\ = > {x}^{2} + \frac{x}{2} - 2 = 0$

Further simplifying,

We can write,

$= > ( {(x)}^{2} + 2 \times x \times \frac{1}{4} + {( \frac{1}{4} )}^{2}) - 2 - {( \frac{1}{4}) }^{2} = 0$

But,

We know that,

$\large\boxed{{a}^{2} + 2ab + {b}^{2} = {(a + b)}^{2}}$

Therefore,

We get,

$= > {(x + \frac{1}{4} )}^{2} - 2 - \frac{1}{16} = 0 \\ \\ = > {(x + \frac{1}{4} )}^{2} - ( \frac{32 + 1}{16} ) = 0 \\ \\ = > {(x + \frac{1}{4} )}^{2} - \frac{33}{16} = 0 \\ \\ = > {(x + \frac{1}{4} )}^{2} = \frac{33}{16} \\ \\ = > {(x + \frac{1}{4}) }^{2} = {( \pm \frac{ \sqrt{33} }{4} )}^{2}$

Again,

Simplifying the terms,

We get,

$= > x + \frac{1}{4} = \pm \frac{ \sqrt{33} }{4} \\ \\ = > x = - \frac{1}{4} \pm \frac{ \sqrt{33} }{4} \\ \\ = > x = \frac{ - 1 \pm \sqrt{33} }{4}$

Hence,

The required roots of the Equation are,

$\large\bold{x = \frac{ - 1 + \sqrt{33} }{4}} \: and \: \large\bold{ x = \frac{ - 1 - \sqrt{33} }{4} }$