How to show that the equation ax^2-(a+b)x+b=0 has a solution for all values of a and b?

2 Answers
Jan 24, 2018

See the explanation below

Explanation:

The quadratic equation is

ax^2-(a+b)x+b=0

In order for this quadratic equation to have solutions, the discriminant >=0

The discriminant is

Delta=(-(a+b))^2-4*(a)*(b)

=(a+b)^2-4ab

=a^2+2ab+b*2-4ab

=a^2-2ab+b^2

=(a-b)^2

AA a,b in RR^2, =>, Delta>=0

The roots are

x=((a+b)+-sqrt((a-b)^2))/(2a)

x=(a+b)+-(a-b)/(2a)

x_1=(a+b+a-b)/(2a)=1

x_2=(a+b-a+b)/(2a)=b/a

The roots are =1 and b/a

Therefore, whatever values of a!=0 " and "b, the quadratic equation has a solution.

Jan 24, 2018

Please see be,ow.

Explanation:

ax^2-(a+b)x+b can be factored to get

ax^2-(a+b)x+b = (ax-b)(x-1) = 0

So, x=1 is always a solution.

Also, if a != 0, then x = b/a is a solution as well.