The number of the real roots of the equation | vedclass

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(D) Case-$I$: $x \leq 5$$(x+1)^{2} - (x-5) = \frac{27}{4}$$(x+1)^{2} - (x+1) + 6 = \frac{27}{4}$$(x+1)^{2} - (x+1) - \frac{3}{4} = 0$Let $y = x+1$,the

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$2$

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(D) Case-$I$: $x \leq 5$$(x+1)^{2} - (x-5) = \frac{27}{4}$$(x+1)^{2} - (x+1) + 6 = \frac{27}{4}$$(x+1)^{2} - (x+1) - \frac{3}{4} = 0$Let $y = x+1$,then $y^{2} - y - \frac{3}{4} = 0 \implies 4y^{2} - 4y - 3 = 0$$(2y-3)(2y+1) = 0 \implies y = \frac{3}{2}, -\frac{1}{2}$$x+1 = \frac{3}{2} \implies x = \frac{1}{2}$ (Valid as $\frac{1}{2} \leq 5$)$x+1 = -\frac{1}{2} \implies x = -\frac{3}{2}$ (Valid as $-\frac{3}{2} \leq 5$)Case-$II$: $x > 5$$(x+1)^{2} + (x-5) = \frac{27}{4}$$x^{2} + 2x + 1 + x - 5 = \frac{27}{4}$$x^{2} + 3x - 4 = \frac{27}{4} \implies 4x^{2} + 12x - 16 = 27$$4x^{2} + 12x - 43 = 0$$x = \frac{-12 \pm \sqrt{144 - 4(4)(-43)}}{8} = \frac{-12 \pm \sqrt{144 + 688}}{8} = \frac{-12 \pm \sqrt{832}}{8}$Since $\sqrt{832} \approx 28.8$,$x = \frac{-12 + 28.8}{8} \approx 2.1$ (Rejected as $x > 5$)Thus,there are $2$ real roots.

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

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