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

Question

Accepted Answer

"When $x > 5$

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

$\Rightarrow {x^2} + 3x - 4 = {{27} \over 4}$

$\Rightarrow {x^2} + 3x - {{43} \over 4} = 0$

$\Rightarrow 4{x^2} + 12x - 43 = 0$

$x = {{ - 12 \pm \sqrt {144 + 688} } \over 8}$

$x = {{ - 12 \pm \sqrt {832} } \over 8} = {{ - 12 \pm 28.8} \over 8}$

$= {{ - 3 + 7.2} \over 2}$

$= {{ - 3 + 7.2} \over 2},{{ - 3 - 7.2} \over 2}$

= 2.1, -5.1 [ both are rejected as x should be > 5 ]

(Therefore no solution)

For $x \le 5$

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

${x^2} + x + 6 - {{27} \over 4} = 0$

$4{x^2} + 4x - 3 = 0$

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

$x = {{ - 4 \pm 8} \over 8} \Rightarrow x = - {{12} \over 8},{4 \over 8}$

$\therefore$ So, the equation have two real roots."

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

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

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