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

A quadratic equation is in the form ax^{2} + bx + c = 0, where a ≠ 0.

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

Let's see the detailed solution.

**Explanation:**

-1 and - b would satisfy the quadratic equation if they are the roots of the quadratic equation.

On 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, where b ≠ 0

therefore, k - 1 = 0

⇒ k = 1

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

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