Question

solve the equation 4x²+3x+5 by completing square method​

Answer 1

Answer:

x does not have any real value for the equation but it has imagary value.

Step-by-step explanation:

Given

$\large p(x)= 4x^2+3x+5$

We have to find  by completing square method​.

First divide the whole equation by coefficient of x^2 ( here 4)

$\large p(x)= \dfrac{4x^2}{4} +\dfrac{3x}{4} +\dfrac{5}{4}=0 \\\\\\\large p(x)=x^2 +\dfrac{3x}{4} +\dfrac{5}{4}=0$

$\large \text{Transfer \dfrac{5}{4} \ to \ RHS \ side }\\\\\\\\\\\large \text{And adding (\dfrac{1}{2})\times \ coefficient \ of \ x }\\\\\\\\\large x^2+\dfrac{3x}{4}+ (\dfrac{1}{2}\times\dfrac{3}{4})^2=-\dfrac{5}{4}+(\dfrac{1}{2}\times\dfrac{3}{4})^2$

$\large x^2+\dfrac{3x}{4}+ (\dfrac{3}{8})^2=-\dfrac{5}{4}+(\dfrac{9}{64})\\\\\\\\\large x^2+\dfrac{3x}{4}+ (\dfrac{3}{8})^2= (\dfrac{-80+9}{64})\\\\\\\\\large x^2+\dfrac{3x}{4}+ (\dfrac{3}{8})^2= (\dfrac{-71}{64})\\\\\\\large x^2+\dfrac{3x}{4}+ (\dfrac{3}{8})^2=(x+\dfrac{3}{8})^2\\\\\\\\\large (x+\dfrac{3}{8})^2= (\dfrac{-71}{64})$

$\large \text{ (x+\dfrac{3}{8})=\sqrt{(\dfrac{-71}{64})} }\\\\\\\\\large \text{ (x+\dfrac{3}{8})= (\dfrac{\sqrt{-71}}{8}) }\\\\\\\\\large \text{x does not have any real value for the equation but it has imagary value.}\\\\\\\\\large \text{ x=\dfrac{\sqrt{-71}}{8}-\dfrac{3}{8}=\dfrac{\sqrt{-71}-3}{8} }$