The equation -x-3-4 -x-1-x-8-6 -x-1-1 has

The correct option is C more than two solutionsPut √(x 1) = t ⇒ x 1 = t^2 or x=t^2 + 1 √(t^2 + 1 + 3 4t) + √(t^2 + 1 + 8 6t) = 1 where t ≥ 0 ⇒ |t 2|+|t 3 ...

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Standard XII

Mathematics

Roots of a Quadratic Equation

The equation ...

Question

The equation $\sqrt{x + 3 - 4 \sqrt{x - 1}} + \sqrt{x + 8 - 6 \sqrt{x - 1}} = 1$ has

no solution

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one solution

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two solutions

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The correct option is D more than two solutions
$P u t \sqrt{x - 1} = t \Rightarrow x - 1 = t^{2} o r x = t^{2} + 1 \sqrt{t^{2} + 1 + 3 - 4 t} + \sqrt{t^{2} + 1 + 8 - 6 t} = 1 w h e r e t \geq 0 \Rightarrow | t - 2 | + | t - 3 | = 1, w h e r e t \geq 0$ . This equation will be satisfied, if $2 \leq t \leq 3$
Therefore, $2 \leq \sqrt{x - 1} \leq 3 o r 5 \leq x \leq 10$ $∴$ The given equation is satisfied for all values of lying in [5, 10]

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The equation √x+3−4√x−1+√x+8−6√x−1=1 has

The equation $\sqrt{x + 3 - 4 \sqrt{x - 1}} + \sqrt{x + 8 - 6 \sqrt{x - 1}} = 1$ has