Question

The roots of (x-a)(x-b)=b2 are​

Answer 1

Answer:

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Answer 2

Given: The equation as: ( x - a ) ( x - b ) = b^2

To find: The roots of the given equation?

Solution:

• Now we have given the equation as:

             ( x - a ) ( x - b ) = b^2

• Now expanding it, we get:

             x^2 - ax - bx + ab = b^2

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

• Now by the roots formula, we have:

             - (-(a+b)) ± √((-(a+b))^2 - 4(ab - b^2)) / 2

             a+b ± √(a^2 + b^2 + 2ab - 4ab + 4b^2) / 2

             a+b ± √((a-b)^2 + 4b^2) / 2

• Now here Discriminant should be greater than or equal to 0.

             ((a-b)^2 + 4b^2) > 0

             (a-b)^2 + 4b^2 > 0

• So here exist two distinct real roots of the equation given.

             a+b + √((a-b)^2 + 4b^2) / 2 and a+b - √((a-b)^2 + 4b^2) / 2

Answer:

            So the roots are:  

            a+b + √((a-b)^2 + 4b^2) / 2 and a+b - √((a-b)^2 + 4b^2) / 2