The number of the real roots of the equation x -12- x -5-27-4 is

For x≥ 5, (x+1)^2+(x 5)=274 ⇒ x^2+3x 4=274 ⇒ x^2+3x 434=0 ⇒ 4x^2 +12x 33=0 x= 12±√(144+688)8 x= 12±√(832)8= 12±28.88 = 3± 7.22 = 3+7.22, 3 7.22 nbsp; nbsp; ...

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Standard XIII

Mathematics

Quadratic Formula for Finding Roots

The number of...

Question

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

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Solution

For $x \geq 5,$
$(x + 1)^{2} + (x - 5) = \frac{27}{4}$
$\Rightarrow x^{2} + 3 x - 4 = \frac{27}{4}$
$\Rightarrow x^{2} + 3 x - \frac{43}{4} = 0$
$\Rightarrow 4 x^{2} + 12 x - 33 = 0$
$x = \frac{- 12 \pm \sqrt{144 + 688}}{8}$
$x = \frac{- 12 \pm \sqrt{832}}{8} = \frac{- 12 \pm 28.8}{8}$
$= \frac{- 3 \pm 7.2}{2}$
$= \frac{- 3 + 7.2}{2}, \frac{- 3 - 7.2}{2}$ (therefore, no solution)

For $x < 5,$
$(x + 1)^{2} - (x - 5) = \frac{27}{4}$
$\Rightarrow x^{2} + x + 6 - \frac{27}{4} = 0$
$\Rightarrow 4 x^{2} + 4 x - 3 = 0$
$x = \frac{- 4 \pm \sqrt{16 + 48}}{8}$
$x = \frac{- 4 \pm 8}{8}$
$\Rightarrow x = - \frac{12}{8}, \frac{4}{8}$
$∴ 2$ real roots.

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The number of the real roots of the equation (x+1)2+|x−5|=274 is

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