Question

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

Answer 1

$\underline{\textsf{Given:}}$

$\textsf{-5 is a root of the equation}\;\mathsf{2x^2+px-15=0}$

$\textsf{and}\;\mathsf{p(x^2+x)+k=0}\;\textsf{has equal roots}$

$\underline{\textsf{To find:}}$

$\textsf{The value of p and k}$

$\underline{\textsf{Solution:}}$

$\textsf{Let the other root of}$

$\mathsf{2x^2+px-15=0}\;\textsf{be 'u'}$

$\bf\textsf{Product of the roots:}$

$\mathsf{u(-5)=\dfrac{-15}{2}}$

$\mathsf{u=\dfrac{3}{2}}$

$\bf\textsf{Sum of the roots:}$

$\mathsf{\dfrac{3}{2}+(-5)=\dfrac{-p}{2}}$

$\mathsf{\dfrac{3-10}{2}=\dfrac{-p}{2}}$

$\mathsf{\dfrac{-7}{2}=\dfrac{-p}{2}}$

$\implies\boxed{\mathsf{p=7}}$

$\textsf{Since}\;\mathsf{px^2+px+k=0}\textsf{has equal roots, we have}$

$\mathsf{b^2-4ac=0}$

$\mathsf{p^2-4(p)(k)=0}$

$\mathsf{7^2-4(7)(k)=0}$

$\mathsf{49-28\,k=0}$

$\mathsf{49=28\,k}$

$\mathsf{k=\dfrac{49}{28}}$

$\implies\boxed{\mathsf{k=\dfrac{7}{4}}}$

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a) 2x^2– 9x + 13 = 0

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