The number of the real roots of the equation x + 1 2 + | x - 5 | = 27 4 is ________.

Case- I x ≤ 5 x + 1 2 - x - 5 = 27 4 x + 1 2 - x + 1 - 3 4 = 0 x + 1 = 3 2 , - 1 2 x = 1 2 , - 3 2 Case- II x > 5 x + 1 + x - 5 = 27 4 x + 1 2 + x + 1 - 51 4 = 0 x = - 1 ± 52 2 (rejected as x > 5 ) So, the equation have two real root.

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Mathematics - Quadratic Equation Question with Solution | TestHub

MathematicsQuadratic EquationTheory of equationsEasy2 minPYQ_2021

MathematicsEasynumerical

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

Answer:

2.00

Solution:

Case- $I$

$x \leq 5$

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

${(x + 1)}^{2} - (x + 1) - \frac{3}{4} = 0$

$x + 1 = \frac{3}{2}, - \frac{1}{2}$

$x = \frac{1}{2}, - \frac{3}{2}$

Case- $II$

$x > 5$

$(x + 1) + (x - 5) = \frac{27}{4}$

${(x + 1)}^{2} + (x + 1) - \frac{51}{4} = 0$

$x = \frac{- 1 \pm \sqrt{52}}{2}$ (rejected as $x > 5$ )

So, the equation have two real root.

Stream:JEESubject:MathematicsTopic:Quadratic EquationSubtopic:Theory of equations

⏱ 2mℹ️ Source: PYQ_2021

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