Question

2xsquare +x-4=0 by completing square method ​

Answer 1

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

• equation :- 2x² + x - 4 = 0

$\large\bf\underline {To \: find:-}$

• Find the value of x by completing the square method

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

equation :- 2x² + x - 4 = 0

Divide both side by 2.

$\rightarrowtail \rm \: 2 {x}^{2} + x - 4 = 0 \\ \\ \rightarrowtail \rm \: \frac{ {2x}^{2} }{2} + \frac{x}{2} - \frac{4}{2} = \frac{0}{2} \\ \\ \rightarrowtail \rm \: {x}^{2} + \frac{x}{2} - 2 = 0$

write the equation in the form :-

$\: \: \: \: \: \: \blacktriangleright \: \: \: \: \: \: \: { \rm {x}^{2} + bx = c}$

$\rightarrowtail \rm \: {x}^{2} + \frac{x}{2} = 2$

Add (1/4)² on both sides.

$\rightarrowtail \rm \: {x}^{2} + \frac{x}{2} + \frac{1}{16} = 2 + \frac{1}{16} \\ \\ \rightarrowtail \rm \: (x + \frac{1}{4} ) {}^{2} = 2 + \frac{1}{16} \\ \\ \rightarrowtail \rm \: (x + \frac{1}{4} {)}^{2} = \frac{32 + 1}{16} \\ \\ \rightarrowtail \rm \: (x + \frac{1}{4} {)}^{2} = \frac{33}{16} \\ \\ \rightarrowtail \rm \: x + \frac{1}{4} = \sqrt{ \frac{33}{16} } \\ \\ \rightarrowtail \rm \: x + \frac{1}{4} = \frac{ \sqrt{33} }{4} \\ \\ \rightarrowtail \rm \: x = \frac { \pm\sqrt{33} }{4} - \frac{1}{4} \\ \\ \rightarrowtail \rm \: x = \frac{ \sqrt{33} - 1}{4} \: \: \\ \\ \rightarrowtail \rm \: or \: \: x = \frac{ - ( \sqrt{33} + 1) }{4}$

So, the roots of the given equation are :-

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

Answer 2

$\sf\blue{Correct \ equation:}$

$\sf{Find \ the \ roots \ of \ the \ quadratic \ equation}$

$\sf{2x^{2}+x-4=0 \ by \ completing \ the}$

$\sf{square \ method.}$

____________________________________

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{\frac{-1+\sqrt33}{4} \ and \ \frac{-1-\sqrt33}{4} \ are \ roots \ of \ the}$

$\sf{given \ quadratic \ equation.}$

$\sf\orange{Given:}$

$\sf{The \ given \ quadratic \ equation \ is}$

$\sf{\implies{2x^{2}+x-4=0}}$

$\sf\pink{To \ find:}$

$\sf{The \ roots \ of \ the \ equation.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{The \ given \ quadratic \ equation \ is}$

$\sf{\implies{2x^{2}+x-4=0}}$

$\sf{Dividing \ the \ equation \ throughout \ by \ 2}$

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

$\sf{\implies{x^{2}+\frac{x}{2}=2}}$

__________________________________

$\sf{(\frac{1}{2}\times \ Coefficient \ of \ x)^{2}}$

$\sf{\implies{(\frac{1}{2}\times\frac{1}{2})^{2}}}$

$\sf{\implies{(\frac{1}{4})^{2}}}$

$\sf{\implies{\frac{1}{16}}}$

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$\sf{Add \ \frac{1}{16} \ on \ both \ sides \ of \ the \ equation}$

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

$\sf{\implies{x^{2}+\frac{x}{2}+\frac{1}{16}=\frac{32+1}{16}}}$

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

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\sf{\implies{x+\frac{1}{4}=\frac{\sqrt33}{4} \ or \ -\frac{\sqrt33}{4}}}$

$\sf{\implies{x=\frac{-1}{4}+\frac{\sqrt33}{4} \ or \ \frac{-1}{4}-\frac{\sqrt33}{4}}}$

$\sf{\implies{x=\frac{-1+\sqrt33}{4} \ or \ \frac{-1-\sqrt33}{4}}}$

$\sf\purple{\tt{\therefore{\frac{-1+\sqrt33}{4} \ and \ \frac{-1-\sqrt33}{4} \ are \ roots \ of \ the}}}$

$\sf\purple{\tt{given \ quadratic \ equation.}}$