f4 quadratic equation

1)(x-4)(x+5)=(a-4)(a+5)
x^2 - 4x + 5x - 20 = (a-4)(a+5)
x^2 + x - [20+(a-4)(a+5)] = 0
It is easy to know that a is one root of x,
Since sum of the roots = - 1
so the other root = - 1 - a = - a - 1
ANSWER : C
2a ) Find the value of k

x^2+k^2=(2k-5)x
x^2 + (5 - 2k)x + k^2 = 0
△ = (5-2k)^2 - 4k^2 = 0
4k^2 - 20k + 25 - 4k^2 = 0
20k = 25
k = 5/4
(b)Hence solve the equation
x^2 + (5 - 2k)x + k^2 = 0
x = [- (5 - 2k) +/- 『the square root of △ (= 0)』] / 2
So x = (2k-5)/2
x = [2(5/4) - 5]/2
x = - 5/4

(x-4)(x+5)=(a-4)(a+5)

=>x^2+x-20=a^2+a-20

=>x^2+x-(a^2+a)=0

=>x^2+x-(a)(a+1)=0

=>(x-a)[x+(a+1)]=0

=>x=a or x=-(a+1)//

SO THE ANSWER IS C//

2. Suppose the quadratic euation x^2+k^2=(2k-5)x has two equal real roots

=>(2k-5)^2-4(k^2)=0

=>4k^2-20k+25-4k^2=0

=>k=5/4//

Sum of roots = @+@

=>(2k-5)=2@

=>@=(1/2)(2k-5)=(1/2)[2(5/4)-5]=-5/4//