Question

find the roots of the equation
(x+2)^2=4(x+1)-1​

Answer 1

Answer:

hope you got your answer

Answer 2

Answer:

$\sf{The \ roots \ of \ the \ given \ equation \ are}$

$\sf{\sqrt{-1} \ and \ -\sqrt{-1}.}$

Given:

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

$\sf{\longmapsto{(x+2)^{2}=4(x+1)-1}}$

To find:

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

Solution:

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

$\sf{\longmapsto{(x+2)^{2}=4(x+1)-1}}$

$\sf{\leadsto{x^{2}+4x+4=4x+4-1}}$

$\sf{\leadsto{x^{2}+1=0}}$

$\sf{Here, \ a=1, \ b=0 \ and \ c=1}$

$\sf{b^{2}-4ac=0^{2}-4(1)(1)}$

$\sf{\leadsto{b^{2}-4ac=-4}}$

$\sf{By \ quadratic \ formula}$

$\sf{\mapsto{x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}}}$

$\sf{\mapsto{x=\dfrac{-0\pm\sqrt{-4}}{2(1)}}}$

$\sf{\mapsto{x=\dfrac{\pm2\sqrt{-1}}{2}}}$

$\sf{\mapsto{x=\pm\sqrt{-1}}}$

$\sf{\mapsto{x=\sqrt{-1} \ or \ -\sqrt{-1}}}$

$\sf\purple{\tt{\therefore{The \ roots \ of \ the \ given \ equation \ are}}}$

$\sf\purple{\tt{\sqrt{-1} \ and \ -\sqrt{-1}.}}$