If and are the roots of the equation ( - 2 + 3 ) ( x - 3 ) + ( x - 6 x ) + ( 9 - 2 3 ) = 0 , x 0 , then is equal to: [2026]

Question

If $α$ and $β (α < β)$ are the roots of the equation $(- 2 + \sqrt{3}) (| \sqrt{x} - 3 |) + (x - 6 \sqrt{x}) + (9 - 2 \sqrt{3}) = 0, x \geq 0,$ then $\sqrt{\frac{β}{α}} + \sqrt{α β}$ is equal to: [2026]

Smart Achievers · Accepted Answer

(2)

$(x - 6 \sqrt{x} + 9) - (2 - \sqrt{3})$ $| \sqrt{x} - 3 |$ $- 2 \sqrt{3} = 0$

$\Rightarrow$ $| \sqrt{x} - 3 |^{2}$ $- (2 - \sqrt{3})$ $| \sqrt{x} - 3 |$ $- 2 \sqrt{3} = 0$

$\Rightarrow | \sqrt{x} - 3 | = 2 or | \sqrt{x} - 3 | = - \sqrt{3} (not possible)$

$\Rightarrow \sqrt{x} = 1 or 5$

$\Rightarrow x = 1 or 25$

$\Rightarrow α = 1 and β = 25$

Aliter:

Let $x \geq 9$ , let $\sqrt{x} = t \Rightarrow t \geq 3$

$(\sqrt{3} - 2) (t - 3) + {(t - 3)}^{2} - 2 \sqrt{3} = 0$

Let $t - 3 = u$

$u^{2} + (\sqrt{3} - 2) u - 2 \sqrt{3} = 0$

$\Rightarrow u = 2 or u = - \sqrt{3}$

$\Rightarrow t - 3 = 2 or t - 3 = - \sqrt{3}$

$\Rightarrow t = 5 or t = 3 - \sqrt{3} (rejected)$

$\Rightarrow x = 25$

Now let $0 < x < 9$

$- (\sqrt{3} - 2) (t - 3) + {(t - 3)}^{2} - 2 \sqrt{3} = 0$

Let $t - 3 = u$

$u^{2} - (\sqrt{3} - 2) u - 2 \sqrt{3} = 0$

$\Rightarrow u = \sqrt{3} or u = - 2$

$\Rightarrow t = 3 + \sqrt{3} (rejected) or t - 3 = - 2$

$\Rightarrow t = 1 \Rightarrow x = 1$

$α = 1, β = 25$

$Now \sqrt{\frac{β}{α}} + \sqrt{α β} = \sqrt{25} + \sqrt{25} = 10$

Solution

Q.
If $α$ and $β (α < β)$ are the roots of the equation $(- 2 + \sqrt{3}) (| \sqrt{x} - 3 |) + (x - 6 \sqrt{x}) + (9 - 2 \sqrt{3}) = 0, x \geq 0,$ then $\sqrt{\frac{β}{α}} + \sqrt{α β}$ is equal to: [2026]

1 8

2 10

3 9

4 11

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Solution

Q. If α and β(α<β) are the roots of the equation (-2+3)(|x-3|)+(x-6x)+(9-23)=0, x≥0, then βα+αβ is equal to: [2026]

1 8

2 10

3 9

4 11

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Q.
If $α$ and $β (α < β)$ are the roots of the equation $(- 2 + \sqrt{3}) (| \sqrt{x} - 3 |) + (x - 6 \sqrt{x}) + (9 - 2 \sqrt{3}) = 0, x \geq 0,$ then $\sqrt{\frac{β}{α}} + \sqrt{α β}$ is equal to: [2026]