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Quadratic Equation and Inequalities question

2021 · 24 Feb · Shift 2 · Q39
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  5. /2021 · 24 Feb · Shift 2 · Q39

Quadratic Equation and Inequalities question

2021 · 24 Feb · Shift 2 · Q39

JEE MainMathematicsQuadratic Equation and InequalitiesNumerical+4 / −1
The number of the real roots of the equation (x+1)2+∣x−5∣=274{(x + 1)^2} + |x - 5| = {{27} \over 4}(x+1)2+∣x−5∣=427​ is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 2

We need to find the number of real roots of

(x+1)2+∣x−5∣=274.(x+1)^2+|x-5|=\frac{27}{4}.(x+1)2+∣x−5∣=427​.

1. Split using the modulus

Since

∣x−5∣={x−5,x≥5,5−x,x<5,|x-5|= \begin{cases} x-5, & x\ge 5,\\ 5-x, & x<5, \end{cases}∣x−5∣={x−5,5−x,​x≥5,x<5,​

we solve in two cases.

2. Case I: x≥5x\ge 5x≥5

Then

(x+1)2+(x−5)=274.(x+1)^2+(x-5)=\frac{27}{4}.(x+1)2+(x−5)=427​.

Expand:

x2+2x+1+x−5=274x^2+2x+1+x-5=\frac{27}{4}x2+2x+1+x−5=427​ x2+3x−4=274.x^2+3x-4=\frac{27}{4}.x2+3x−4=427​.

Bring all terms to one side:

x2+3x−434=0.x^2+3x-\frac{43}{4}=0.x2+3x−443​=0.

Multiply by 444:

4x2+12x−43=0.4x^2+12x-43=0.4x2+12x−43=0.

Now solve:

x=−12±122−4⋅4⋅(−43)8=−12±144+6888=−12±8328.x=\frac{-12\pm\sqrt{12^2-4\cdot 4\cdot(-43)}}{8} =\frac{-12\pm\sqrt{144+688}}{8} =\frac{-12\pm\sqrt{832}}{8}.x=8−12±122−4⋅4⋅(−43)​​=8−12±144+688​​=8−12±832​​.

Since

832=813,\sqrt{832}=8\sqrt{13},832​=813​,

we get

x=−12±8138=−3±2132.x=\frac{-12\pm 8\sqrt{13}}{8}=\frac{-3\pm 2\sqrt{13}}{2}.x=8−12±813​​=2−3±213​​.

Check which satisfy x≥5x\ge 5x≥5:

  • −3+2132≈−3+7.212≈2.1<5\frac{-3+2\sqrt{13}}{2}\approx \frac{-3+7.21}{2}\approx 2.1<52−3+213​​≈2−3+7.21​≈2.1<5
  • −3−2132<0\frac{-3-2\sqrt{13}}{2}<02−3−213​​<0

So no root from this case.

3. Case II: x<5x<5x<5

Then

(x+1)2+(5−x)=274.(x+1)^2+(5-x)=\frac{27}{4}.(x+1)2+(5−x)=427​.

Expand:

x2+2x+1+5−x=274x^2+2x+1+5-x=\frac{27}{4}x2+2x+1+5−x=427​ x2+x+6=274.x^2+x+6=\frac{27}{4}.x2+x+6=427​.

So

x2+x−34=0.x^2+x-\frac{3}{4}=0.x2+x−43​=0.

Multiply by 444:

4x2+4x−3=0.4x^2+4x-3=0.4x2+4x−3=0.

Factorize:

4x2+4x−3=(2x+3)(2x−1)=0.4x^2+4x-3=(2x+3)(2x-1)=0.4x2+4x−3=(2x+3)(2x−1)=0.

Hence

x=−32,x=12.x=-\frac{3}{2},\quad x=\frac{1}{2}.x=−23​,x=21​.

Both satisfy x<5x<5x<5.

So this case gives 2 real roots.

4. Total number of real roots

Thus the equation has

222

real roots.

5. Comparison with stored answer

Stored correct answer: 222

Our derived answer matches the stored answer.

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