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If −5 is a Root of the Quadratic Equation 2 X 2 + P X − 15 = 0 and the Quadratic Equation P ( X 2 + X ) + K = 0 Has Equal Roots, Find the Value of K. - CBSE Class 10 - Mathematics

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Question

If −5 is a root of the quadratic equation\[2 x^2 + px - 15 = 0\] and the quadratic equation \[p( x^2 + x) + k = 0\] has equal roots, find the value of k.

Solution

The given quadratic equation is \[2 x^2 + px - 15 = 0\] and one root is −5.
Then, it satisfies the given equation.

\[2 \left( - 5 \right)^2 + p\left( - 5 \right) - 15 = 0\]

\[ \Rightarrow 50 - 5p - 15 = 0\]

\[ \Rightarrow - 5p = - 35\]

\[ \Rightarrow p = 7\]

The quadratic equation \[p( x^2 + x) + k = 0\], has equal roots.

Putting the value of p, we get

\[7\left( x^2 + x \right) + k = 0\]

\[ \Rightarrow 7 x^2 + 7x + k = 0\]

Here, 

\[a = 7, b = 7 \text { and } c = k\].

As we know that \[D = b^2 - 4ac\]

Putting the values of  \[a = 7, b = 7 \text { and } c = k\]

\[D = \left( 7 \right)^2 - 4\left( 7 \right)\left( k \right)\]

\[ = 49 - 28k\]

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

Thus, 

\[49 - 28k = 0\]

\[\Rightarrow 28k = 49\]

\[ \Rightarrow k = \frac{49}{28}\]

\[ \Rightarrow k = \frac{7}{4}\]

Therefore, the value of k is  \[\frac{7}{4}\].

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Solution If −5 is a Root of the Quadratic Equation 2 X 2 + P X − 15 = 0 and the Quadratic Equation P ( X 2 + X ) + K = 0 Has Equal Roots, Find the Value of K. Concept: Solutions of Quadratic Equations by Factorization.
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