Number of real roots of equation (x+1)^{2}+|x-5|=frac{27}{4} is ....... .

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4-2.Quadratic Equations and Inequations

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The number of the real roots of the equation $(x+1)^{2}+|x-5|=\frac{27}{4}$ is ....... .

$6$

$0$

$4$

$2$

(JEE MAIN-2021)

Case$-I$

$x \leq 5$

$(x+1)^{2}-(x-5)=\frac{27}{4}$

$(x+1)^{2}-(x+1)-\frac{3}{4}=0$

$x+1=\frac{3}{2},-\frac{1}{2}$

$x=\frac{1}{2},-\frac{3}{2}$

Case$-II$

$x >5$

$(x+1)+(x-5)=\frac{27}{4}$

$(x+1)^{2}+(x+1)-\frac{51}{4}=0$

$x=\frac{-1 \pm \sqrt{52}}{2}($ rejected as $x > 5)$

So, the equation have two real root.

Standard 11

Mathematics

normal

(KVPY-2017)

hard

normal

hard

(JEE MAIN-2024)

normal

(KVPY-2015)