Question

find the values of a for which one root of the equation (a - 5) x ^ 2 - 2ax + a - 4 = 0 a smaller than an other greater than 2

Answer 1

Answer:

Given equation is (a−5)x 2 −2ax+a−4=0. It has one root smaller than 1 and other greater than 2.

Then the product of roots will be less than 2. So putting values we get,

(a−5)/(a−4) <2⇒a>6.....(1)

Now since root is greater than 2, then the function should have a negative value at a value of 2.

This means, (a−5)2^2 −2a×2+a−4<0⇒a<24....(2)

Now we will check the roots for a=6 and a=24

1. For a=6, we have x^2 −12x+2=0⇒x=6±√34

So, we can see that one of the roots here is less than 1 and other is greater than 2. So a=6 is accepted value.

2. For a=24, we get 19x^2−48x+20=0⇒x=2,0

In the obtained roots, one root is equal to 2, so a=24 is not a acceptable value.

So, then acceptable integral values of a will be 6,7,8,9,... up to 23.

Sum of these values =261