If the roots of the quadratic equation kx² + (a + b)x + ab are (-1) and (-b), then the value of k is?

An equation of the form ax² + bx + c = 0, where a ≠ 0 is called a quadratic equation.

Answer: The value of k is 1, if the roots of the quadratic equation kx² + (a + b)x + ab are (-1) and (-b).

Let's find the value of k

Explanation:

Since '-1' and '-b' are the roots of the given quadratic equation this would mean that they would satisy the given quadratic equation.

Substituting x = -b in the quadratic equation we get,

k(-b)² + (a + b)(-b) + ab = 0

kb² - ab - b² + ab = 0

kb² - b² = 0

b²(k - 1) = 0

Since b² cannot be equal to 0

therefore, k - 1 = 0

k = 1

Thus, the value of k is 1.