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# If the roots of the quadratic equation kx^{2} + (a + b)x + ab are (-1) and (-b), then the value of k is?

An equation of the form ax^{2} + 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 kx^{2} + (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

kb^{2 }- ab - b^{2 }+ ab = 0

kb^{2 }- b^{2 }= 0

b^{2}(k - 1) = 0

Since b^{2} cannot be equal to 0

therefore, k - 1 = 0

k = 1

### Thus, the value of k is 1.

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