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If the Sum of the Roots of the Equation X 2 − ( K + 6 ) X + 2 ( 2 K − 1 ) = 0 is Equal to Half of Their Product, Then K = - CBSE Class 10 - Mathematics

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Question

If the sum of the roots of the equation \[x^2 - \left( k + 6 \right)x + 2\left( 2k - 1 \right) = 0\] is equal to half of their product, then k =

  • 6

  • 7

  • 1

  • 5

Solution

The given quadric equation is `x^2 - (k+6)x + 2 (2k - 1) = 0 `, and roots are equal

Then find the value of k.

Let `alpha and beta ` be two roots of given equation

And, a = 1, b = -(k + 6) and , c = 2 (2k - 1)

Then, as we know that sum of the roots

`alpha + beta = (-b)/a`

`alpha + beta = (-{-(k + 6)})/1`

          `= (k + 6)`

And the product of the roots

     `alpha . beta = c /a`

        `alphabeta = (2(2k - 1))/1`

             ` = 2 (2k- 1)`

According to question, sum of the roots ` = 1/2 xx` product of the roots

`(k + 6) = 1/2 xx 2 (2k - 1)`

   `k+6 = 2k - 1`

    ` 6+1 = 2k - k`

            7 = k

Therefore, the value of k = 7.

  Is there an error in this question or solution?

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Solution If the Sum of the Roots of the Equation X 2 − ( K + 6 ) X + 2 ( 2 K − 1 ) = 0 is Equal to Half of Their Product, Then K = Concept: Solutions of Quadratic Equations by Factorization.
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