If $- 5$ is one of the zeroes of $2 x^{2} + p x - 15$ . Quadratic polynomial $p (x^{2} + x) + k$ has both the zeros equal to each other. Then find $k$ .

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Solution

Given, $2 x^{2} + p x - 15 = 0$

As $- 5$ is one of the zeroes of the above equation.

$⟹ 2 (- 5)^{2} + p (- 5) - 15 = 0$

$⟹ 50 - 5 p - 15 = 0$

$⟹ 5 p = 35$

$⟹ p = 7$

Putting the value of $p$ in the given polynomial we get

$7 x^{2} + 7 x + k$

It is given that it has both the zeros equal to each other

$∴ D = 0 ⟹ b^{2} - 4 a c = 0$

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

$28 k = 49$

$k = \frac{7}{4}$

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If −5 is one of the zeroes of 2x2+px−15. Quadratic polynomial p(x2+x)+k has both the zeros equal to each other. Then find k.

Given, 2x2+px−15=0As −5 is one of the zeroes of the above equation.⟹2(−5)2+p(−5)−15=0⟹50−5p−15=0⟹5p=35⟹p=7Putting the value of p in the given polynomial we get7x2+7x+kIt is given that it has both the zeros equal to each other∴D=0⟹b2−4ac=0(7)2−4(7)(k)=028k=49k=74

If $- 5$ is one of the zeroes of $2 x^{2} + p x - 15$ . Quadratic polynomial $p (x^{2} + x) + k$ has both the zeros equal to each other. Then find $k$ .