Question

Suppose that the two roots of the equation
$\scriptsize\sf \dfrac{1}{x^2 - 10x - 29} + \dfrac{1}{x^2 - 10x - 45} - \dfrac{2}{x^2 - 10x - 69 } = 0$ are $\sf \alpha$ and $\sf \beta.$ Find the value of $\sf (- \alpha \beta).$​

Answer 1

Given : 1/(x^2-10-29) +1/(x^2-10x-45) -2/(x^2-10x-69) = 0

roots are alpha beta

To find : value of -alpha.beta

Solution:

assume x^2-10x = a

1/(a-29) + 1/(a-45) - 2/(a-69) = 0

(2a -74)(a-29)(a -45) = 2/(a-69)

2(a -37)/(a - 29)(a-45) = 2/(a -69)

(a-37)(a-69) = (a -29)(a -45)

-106a + 37×69 = -74a + 29×45

32a = 1248

a = 39

x^2- 10x = 39

x^2 -13x + 3x -39 = 0

(x-13)(x +3) = 0

x = 13 , -3

-(13)(-3) = 39

Answer is 39

Learn more:

For what values of m roots of the quadratic equations will be same1 ...

https://brainly.in/question/9878645