Basics & LogarithmsmediumPYQ · JEE Main · 24 Feb 2021 · Shift 2 (Afternoon)Free

Basics & Logarithms: Number Real Roots Equation (JEE Main 2021)

JEE Maths question with a full step-by-step solution.

Question

The number of real roots of the equation

(x+1)^{2}+|x-5|=\dfrac{27}{4}

Solution

Answer: 2

Step 1: Case I:

x\ge 5

. Then

|x-5|=x-5

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

x^{2}+2x+1+x-5=\dfrac{27}{4}

x^{2}+3x-4=\dfrac{27}{4}

4x^{2}+12x-16=27\;\Rightarrow\;4x^{2}+12x-43=0

Step 2: Solve:

x=\dfrac{-12\pm\sqrt{144+688}}{8}=\dfrac{-12\pm\sqrt{832}}{8}=\dfrac{-12\pm 4\sqrt{52}}{8}=\dfrac{-3\pm 2\sqrt{13}}{2}

\dfrac{-3+2\sqrt{13}}{2}\approx\dfrac{-3+7.2}{2}\approx 2.1

(not

\ge 5

, rejected).

\dfrac{-3-2\sqrt{13}}{2}<0

(rejected). Step 3: Case II:

x<5

. Then

|x-5|=5-x

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

x^{2}+2x+1+5-x=\dfrac{27}{4}

x^{2}+x+6=\dfrac{27}{4}

4x^{2}+4x+24=27\;\Rightarrow\;4x^{2}+4x-3=0

Step 4: Solve:

x=\dfrac{-4\pm\sqrt{16+48}}{8}=\dfrac{-4\pm 8}{8}

x=\dfrac{1}{2}\text{ or }x=-\dfrac{3}{2}

Both satisfy

x<5

. Accepted. Step 5: Total real roots:

2

. Answer:

2

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