JEE PYQ: Quadratic Equation Question 8
Question 8 - 2021 (24 Feb Shift 2)
The number of the real roots of the equation $(x + 1)^2 + |x - 5| = \frac{27}{4}$ is
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Answer: (2)
Solution
For $x \geq 5$:
$(x+1)^2 + (x-5) = \frac{27}{4}$
$\Rightarrow x^2 + 3x - 4 = \frac{27}{4}$
$\Rightarrow x^2 + 3x - \frac{43}{4} = 0$
$\Rightarrow 4x^2 + 12x - 43 = 0$
No real solution in $x \geq 5$.
For $x \leq 5$:
$(x+1)^2 - (x-5) = \frac{27}{4}$
$4x^2 + 4x - 3 = 0$
$\Rightarrow x = \frac{-4 \pm 8}{8} \Rightarrow x = \frac{-12}{8}, \frac{4}{8}$
$\therefore$ 2 Real Roots
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