If α and β(α<β) are the roots of the equation (-2+√3)(|√x-3|)+(x-6 √x)+(9-2 √3)=0, x geqslant 0, then √ β α+√α β is equal to :

Let $t = \sqrt{x} \geq 0$t=x​≥0. The equation becomes $(-2+\sqrt{3})|t-3| + (t^2 - 6t) + (9 - 2\sqrt{3}) = 0$(−2+3​)∣t−3∣+(t2−6t)+(9−23​)=0.

Since $t^2 - 6t + 9 = (t-3)^2$t2−6t+9=(t−3)2, we get $(t-3)^2 + (\sqrt{3}-2)|t-3| - 2\sqrt{3} = 0$(t−3)2+(3​−2)∣t−3∣−23​=0.

Let $u = |t-3| \geq 0$u=∣t−3∣≥0: $u^2 + (\sqrt{3}-2)u - 2\sqrt{3} = 0$u2+(3​−2)u−23​=0.

Discriminant $= (\sqrt{3}-2)^2 + 8\sqrt{3} = 7 + 4\sqrt{3} = (2+\sqrt{3})^2$=(3​−2)2+83​=7+43​=(2+3​)2.

$u = \frac{(2-\sqrt{3}) \pm (2+\sqrt{3})}{2}$u=2(2−3​)±(2+3​)​, giving $u = 2$u=2 or $u = -\sqrt{3}$u=−3​ (rejected).

$|t-3| = 2 \Rightarrow t = 5$∣t−3∣=2⇒t=5 or $t = 1$t=1, so $x = 25$x=25 or $x = 1$x=1.

$\alpha = 1, \beta = 25$α=1,β=25.

$\sqrt{\beta/\alpha} + \sqrt{\alpha\beta} = \sqrt{25} + \sqrt{25} = 5 + 5 = 10$β/α​+αβ​=25​+25​=5+5=10.