Question

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

Find the root by completing the square method.

Answer 1

$\huge \mathfrak \red{answer}$

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Question:

this is the qudratic equation

Find the root by completing the square method.

$\sf \red{ {2x}^{2} + x - 4}$

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

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

$\sf{⇒ {x}^{2} + \frac{x}{2} = 2}$

adding the both sides we get

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

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

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

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

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

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

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hence proved

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do you know what is qudratic equation?

• A qudratic equation is a polynomial equation of second degree

• the general form is

• $\sf{ {ax}^{2} + bx + c = 0}$

• where a, b and c are real number then a is not equal to zero

• A qudratic equation in x is also called a second degree polynomial equation in x

I hope it's help uh

Answer 2

$\large{ \dag{ \bold{ \red{ \underline{ \underline{Question...}}}}}}$

$2 {x}^{2} + x - 4 \\ \\ \sf{ \large{find \: roots \: by \: completing \: square \: method.}}$

$\large{ \dag{ \bold{ \green{ \underline{ \underline{Answer...}}}}}}$

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

$\large{ \dag{ \bold{ \red{ \underline{ \underline{Solution...}}}}}}$

$\sf{ \star{ \purple{given...}}}$

• P(x)= 2x^2+x-4

$\sf{ \star{ \purple{to \: find...}}}$

• Value the roots of p(x)

$\large{ \sf{ \implies{ \: 2 {x}^{2} +x-4 = 0}}} \\ \\ \large{ \sf{ \implies{ \frac{2 {x}^{2} + x - 4 }{2} = \frac{0}{2} }}} \\ \large{ \red{ \sf{dividing \: by \: 2 \: to \: simpify \: coefficient \: f {x}^{2} }}} \\ \\ \large{ \sf{ \implies{ {x}^{2} + \frac{x}{2} -2 = 0}}} \\ \\ \large{ \sf{ \implies{ {x}^{2} + \frac{x}{2} = 2}}} \\ \\ \large{ \sf{ \implies{ {x}^{2} + 2 \times x \times \frac{1}{4} + ( \frac{1}{4} ) {}^{2} = 2 + (\frac{1}{4} ) {}^{2} }}} \\ \\ \large{ \sf{ \implies{(x + \frac{1}{4} ) {}^{2} = 2 + \frac{1}{16} }}} \\ \\ \large{ \sf{ \implies{(x + \frac{1}{4}) {}^{2} = \frac{ 32 + 1}{16} }}} \\ \\ \large{ \sf{ \implies{(x + \frac{1}{4} ) {}^{2} = \frac{ 33}{16} }}} \\ \\ \large{ \sf{ \implies{x + \frac{1}{4} = \pm{ \sqrt{\frac{ 33}{16}} }}}} \\ \\ \\ \large{ \purple{ \boxed{ case - 1)}}} \: \: \large{ \sf{x + \frac{1}{4} = \sqrt{ \frac{33}{16} } }} \\ \\ \large{ \sf{ \implies{ x = \frac{ \sqrt{33} }{4} - \frac{1}{4} }}} \\ \\ \large{ \sf{ \implies{ \red{ \boxed{x = \frac{ \sqrt{33} - 1}{4} }}}}} \\ \\ \large{ \sf{ \purple{ \boxed{case - 2)}}}} \: \: \large{ \sf{x + \frac{1}{4} = - \sqrt{ \frac{33}{16} }}} \\ \\ \large{ \sf{ \implies{x = - \frac{ \sqrt{33} }{4} - \frac{1}{4} }}} \\ \\ \large{ \sf{ \implies{ \red{ \boxed{x = - \frac{( \sqrt{33 } + 1)}{4}}}}}}$

$\large{ \orange{ \underline{ \bold{ \underline{ \star{required \: answer \: is \: \frac{ \sqrt{33} - 1 }{4} and \: \frac{ - ( \sqrt{33} + 1) }{4} }}}}}}$