Question

4x^2+3x+5=0, solve using completing the square root . Class-10

Answer 1

Answer:

4x^2+3x+5=0

(2x)^2+2*2x*3/4+(3/4)^2-(3/4)^2+5=0

(2x+3/4)^2+71/16=0

Answer 2

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

$\sf{Roots \ are \ \frac{-3+\sqrt{-71}}{8} \ and \ \frac{-3-\sqrt{-71}}{8}}$

$\sf\orange{Given:}$

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

$\sf{4x^{2}+3x+5=0}$

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

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

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

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

$\sf{\implies{4x^{2}+3x+5=0}}$

$\sf{Dividing \ the \ equation \ throughout \ by \ 4.}$

$\sf{\implies{x^{2}+\frac{3x}{4}+\frac{5}{4}=0}}$

$\sf{\implies{x^{2}+\frac{3x}{4}=\frac{-5}{4}}}$

_______________________________

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

$\sf{=>(\frac{3}{4}\times\frac{1}{2})^{2}}$

$\sf{=>(\frac{3}{8})^{2}}$

$\sf{=>\frac{9}{64}}$

_______________________________

$\sf{Add \ \frac{9}{64} \ on \ both \ sides \ of \ equation}$

$\sf{\implies{x^{2}+\frac{3x}{4}+\frac{9}{64}=\frac{-5}{4}+\frac{9}{64}}}$

$\sf{\implies{(x+\frac{3}{8})^{2}=\frac{-80+9}{64}}}$

$\sf{\implies{(x+\frac{3}{8})^{2}=\frac{-71}{64}}}$

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

$\sf{\implies{x+\frac{3}{8}=\frac{\sqrt{-71}}{8} \ or \ -\frac{\sqrt{-71}}{8}}}$

$\sf{\implies{x=\frac{-3}{8}+\frac{\sqrt{-71}}{8} \ or \ \frac{-3}{8}-\frac{\sqrt{-71}}{8}}}$

$\sf{\implies{x=\frac{-3+\sqrt{-71}}{8} \ or \ \frac{-3-\sqrt{-71}}{8}}}$

$\sf\purple{\tt{\therefore{Roots \ are \ \frac{-3+\sqrt{-71}}{8} \ and \ \frac{-3-\sqrt{-71}}{8}}}}$