Question

4x^2 - 3x - 5 = 0 completing the square method

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Value\:of\:x=\pm\frac{\sqrt{89}+3}{8}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies 4 {x}^{2} - 3x - 5 = 0 \\ \\ \red{\underline \bold{To \: Find:}}\\ \tt: \implies Value \: of \: x = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies {4x}^{2} - 3x - 5 = 0 \\ \\ \tt \circ \: Dividing \: both \: side \: by \: coefficient \: of \: {x}^{2} \\ \tt: \implies \frac{4 {x}^{2} }{4} - \frac{3x}{4} - \frac{5}{4} = 0 \\ \\ \tt: \implies {x}^{2} - \frac{3x}{4} - \frac{5}{4} = 0 \\ \\ \tt \circ \: Adding \: ( \frac{b}{2a} )^{2} \: both \: side= (\frac{ - \frac{3}{4} }{2 \times 1} )^{2} = \frac{9}{64} \\ \\ \tt: \implies {x}^{2} - \frac{3x}{4} + \frac{9}{64} - \frac{5}{4} = \frac{9}{64} \\ \\ \tt: \implies ({x - \frac{3}{8}) }^{2} = \frac{9}{64} + \frac{5}{4} \\ \\ \tt: \implies {(x - \frac{3}{8} )}^{2} = \frac{9 + 80}{64} \\ \\ \tt: \implies {(x - \frac{3}{8}) }^{2} = \frac{89}{64} \\ \\ \tt: \implies x - \frac{3}{8} = \sqrt{ \frac{89}{64} } \\ \\ \tt: \implies x = \pm\frac{ \sqrt{89} }{8} + \frac{3}{8} \\ \\ \tt: \implies x = \pm\frac{ \sqrt{89} + 3}{8}$

Answer 2

Given polynomial:-

$\displaystyle{4x^2-3x-5=0}$

To find:-

$\textsf{The roots of the given polynomial by "completing the square method".}$

Solution:-

$\textsf{Dividing the equation by 4, we get:}\\\\\displaystyle{x^2-\frac{3x}{4}-\frac{5}{4}=0 }\\\\\displaystyle{\implies x^2-\frac{3x}{4} =\frac{5}{4} }$

$\textsf{Multiplying }(\frac{3}{8} )^2 \textsf{ to both the sides;}\\\\\displaystyle{x^2-\frac{3x}{4}+(\frac{3}{8} )^2=\frac{5}{4} +(\frac{3}{8} )^2}\\\\\displaystyle{\implies x^2(-2*x*\frac{3}{8})+(\frac{3}{8} )^2=\frac{5}{4} +\frac{9}{64} }\\\\\displaystyle{\implies (x-\frac{3}{8} )^2=\±\frac{80+9}{64 } }\\\\\displaystyle{\implies x-\frac{3}{8}=\±\sqrt{\frac{89}{64} } }\\\\\displaystyle{\implies x=\±\frac{\sqrt{89} }{8}+\frac{3}{8} }$

$\displaystyle{\boxed{\implies x=\±\frac{\sqrt{89}+3 }{8} }}$