Question

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

Answer 1

Hint: Here, the equation ${\left( {x + 1} \right)^2} + \left| {x - 5} \right| = \dfrac{{27}}{4}$ is given. Make two cases in first one $x \geqslant 5$ where take modulus $\left| {x - 5} \right|$ as positive and in second case take modulus $\left| {x - 5} \right|$as negative for $x < 5$. After solving a quadratic equation find the roots using $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ and check whether the roots satisfy the case or not. If yes, then the roots are real.

Formula used:
General quadratic equation $a{x^2} + bx + c$ whose roots are $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution: 
Given that,
${\left( {x + 1} \right)^2} + \left| {x - 5} \right| = \dfrac{{27}}{4}$
Case 1 –
For $x \geqslant 5$,
${\left( {x + 1} \right)^2} + \left( {x - 5} \right) = \dfrac{{27}}{4}$
${x^2} + 1 + 2x + x - 5 - \dfrac{{27}}{4} = 0$
\[{x^2} + 3x - \dfrac{{43}}{4} = 0\]
$4{x^2} + 12x - 43 = 0$
$x = \dfrac{{ - \left( {12} \right) \pm \sqrt {{{\left( {12} \right)}^2} - 4\left( 4 \right)\left( { - 43} \right)} }}{{2\left( 4 \right)}}$
$x = \dfrac{{ - 12 \pm \sqrt {832} }}{8}$
$x = \dfrac{{ - 12 \pm 28.8}}{8}$
$x = \dfrac{{ - 12 + 28.8}}{8},\dfrac{{ - 12 - 28.8}}{8}$
$x = 3.36, - 5.1$
The roots are not greater than or equal to $5$
Therefore, required roots are not real.
Case 2 –
For $x < 5$,
${\left( {x + 1} \right)^2} - \left( {x - 5} \right) = \dfrac{{27}}{4}$
${x^2} + 1 + 2x - x + 5 - \dfrac{{27}}{4} = 0$
\[{x^2} + x - \dfrac{3}{4} = 0\]
\[4{x^2} + 4x - 3 = 0\]
\[x = \dfrac{{ - 4 \pm \sqrt {{{\left( 4 \right)}^2} - 4\left( 4 \right)\left( { - 3} \right)} }}{{2\left( 4 \right)}}\]
\[x = \dfrac{{ - 4 \pm \sqrt {64} }}{8}\]
\[x = \dfrac{{ - 4 \pm 8}}{8}\]
\[x = 0.5, - 1.5\]
Hence, the equation ${\left( {x + 1} \right)^2} + \left| {x - 5} \right| = \dfrac{{27}}{4}$ has two real roots i.e., \[0.5, - 1.5\].
Hence, Option (3) is the correct answer.

Note: In this problem, students should note that if the modulus is present in the question then two cases will be there greater than or equal to and less than. Also, the alternative method to solve the quadratic equation is compare the equation with general equation $a{x^2} + bx + c$ and calculate the equation using $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.