मराठी

If X 2 + K ( 4 X + K − 1 ) + 2 = 0 Has Equal Roots, Then K =

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प्रश्न

If \[x^2 + k\left( 4x + k - 1 \right) + 2 = 0\] has equal roots, then k =

 

पर्याय

  • \[- \frac{2}{3}, 1\]

  • \[\frac{2}{3}, - 1\]

  • \[\frac{3}{2}, \frac{1}{3}\]

  • \[- \frac{3}{2}, - \frac{1}{3}\]

MCQ
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उत्तर

The given quadric equation is \[x^2 + k\left( 4x + k - 1 \right) + 2 = 0\], and roots are equal

Then find the value of k.

\[x^2 + k\left( 4x + k - 1 \right) + 2 = 0\]

`x^2 + 4kx + (k^2 - k + 2) = 0`

Here,` a = 1 , b = 4k and , c = k^2 - k + 2`

As we know that ` D = b^2 - 4ac`

Putting the value of ` a = 1 , b = 4k and , c = k^2 - k + 2`

`=(4k)^2 - 4xx 1 xx (k^2 - k + 2)`

`= 16k^2 - 4k^2 + 4k - 8`

`=12k^2 + 4k - 8`

`=4 (3k^2 + k - 2)`

The given equation will have real and distinct roots, if D =0 

           `4(3k^2 + k - 2) = 0`

                 `3k^2 + k - 2 = 0`

      `3k^2 + 3k - 2k - 2 = 0`

  `3k (k+1) - 2(k+ 1) = 0`

(k + 1)(3k - 2) = 0

`(k+ 1) = 0`

`k = -1`

or

`(3k - 2) = 0`

             `k = 2/3`

Therefore, the value of `k =2/3: -1`

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पाठ 4: Quadratic Equations - Exercise 4.15 [पृष्ठ ८४]

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आर.डी. शर्मा Mathematics [English] Class 10
पाठ 4 Quadratic Equations
Exercise 4.15 | Q 20 | पृष्ठ ८४

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