Question

SOLVE FOR X BY COMPLETING THE SQUARE METHOD ;;2x^2+x-4=0

Answer 1

Solution:

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Given & To find:

Solve for x using completing the square method,.

2x² + x - 4 = 0,
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As, we know that,

(a + b)² = a² + 2ab + b²

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We are intended to find the value of x, using this formula,.

=> For doing that, we need to bring this equation to that form (a² + 2ab + b² for some value a & b.,.)

We take an attempt to bring this equation to that form,

=> 2x² + x - 4 = 0,

=> $x^2 + \frac{x}{2} - 2 = 0$ (Dividing both the sides by $\frac{1}{2}$ ),

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

 (Multiplying & dividing the  middle term by 2 to get the term 2ab (here a = x, $b = \frac{1}{2}$))

So, we have the form,

a² + 2ab + c  (for some value c, here it is - 4), we need b² to bring this equation to that form,

By adding $( \frac{1}{4}) ^2$ both the sides ,

we get,

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

It can be, reduced to the form (a + b)²

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

=>  $(x + \frac{1}{4} )^2 = \frac{1}{16} + 2$

=> $(x + \frac{1}{4} )^2 = \frac{1+32}{16}$

=> $(x + \frac{1}{4} )^2 = \frac{33}{16}$

By moving the square from LHS to RHS, we get,

=> $(x + \frac{1}{4} ) = \sqrt{ \frac{33}{16} }$
 
=> $(x + \frac{1}{4} ) = \frac{ \sqrt{33} }{4}$ (or) $(x + \frac{1}{4} ) = \frac{ - \sqrt{33} }{4}$

=> $x = \frac{ \sqrt{33} }{4} - \frac{1}{4}$ (or) $x = \frac{ - \sqrt{33} }{4} - \frac{1}{4}$

=> $x = \frac{ \sqrt{33} - 1 }{4}$ (or) $x = \frac{ -\sqrt{33} - 1}{4}$

It's completed,.

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Answer 2

$x=\frac{-1+\sqrt{33} }{4} \ or \ \frac{-1-\sqrt{33} }{4}$ will be the value of x using completing the square method.

Step-by-step explanation:

Given:

equation= 2x²+x-4=0

To find:

value of x using completing the square method

Solution:

First, we simplify the equation to get x² alone

For that, we divide the equation 2x²+x-4=0 by 2

We get

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

$x^{2} +\frac{x}{2}=2$..................(bringing 2 to the right side)

The third term will be  (1/2 x coefficient of $x$ )²= (1/2 x 1/2)²

                          = (1/4)²

                          = 1/16

Adding $\frac{1}{16}$ to both sides of the equation

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

$(x+\frac{1}{4})^{2}= \frac{32+1}{16}$.......................[(a+b)²= a²+2ab+b²]

$(x+\frac{1}{4})^{2}= \frac{33}{16}$

Taking square roots on both sides

$x+\frac{1}{4} = \±\frac{\sqrt{33} }{4}$

$x= \±\frac{\sqrt{33} }{4}-\frac{1}{4}$

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

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

∴ x= $\frac{-1+\sqrt{33} }{4}, \frac{-1-\sqrt{33} }{4}$ will be the solution of the given equation

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