Question

Pls....answer question 64

Answer 1

Answer:

Step-by-step explanation:

$Given: \frac{x-3}{x-4}+\frac{x-5}{x-6}= \frac{10}{6}$

$=> 6(x - 3)(x-6)+6(x-5)(x-4)=10(x-4)(x-6)$

$=> 6x^2 - 54x + 108 + 6x^2 - 54x + 120 = 10x^2 - 100x + 240$

$=>12x^2 - 108x + 228 = 10x^2 - 100x + 240$

$=>12x^2-8x-12=10x^2$

$=>2x^2-8x-12=0$

On comparing with ax^2 + bx + c = 0, a = 2, b = -8, c = -12.

∴ D= b² - 4ac

     = (-8)^2 - 4(2)(-12)

     = 64 + 96

     = 160.

(i)

$=>x=\frac{-b+\sqrt{D}}{2a}$

$=>\frac{-(-8)+\sqrt{160}}{2(2)}$

$=>\frac{8+\sqrt{160}}{4}$

$=>\frac{8+4\sqrt{10}}{4}$

$=>\frac{4(2+\sqrt{10})}{4}$

$=>2+\sqrt{10}$

(ii)

$=>x=\frac{-b-\sqrt{D}}{2a}$

$=>\frac{-(-8)-\sqrt{160}}{2(2)}$

$=>\frac{8-\sqrt{160}}{4}$

$=>\frac{8-4\sqrt{10}}{4}$

$=>\frac{4(2- \sqrt{10})}{4}$

$=>2-\sqrt{10}$

Therefore, the solutions are:

$=> \boxed{2 +\sqrt{10}, 2-\sqrt{10}}$

Hope it helps!

Answer 2

Bhai sorry but isse jyada ho hi nhi rha hai