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Discriminant

If px2 +qx+...

Question

If $p x^{2} + q x + r = 0$ has equal roots, then $q^{2} =$

A

4 p

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B

4 p r

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C

8 p r

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D

p r

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Solution

The correct option is D $4 p r$
$p x^{2} + q x + r = 0$

The standard quadratic equation is $a x^{2} + b x + c = 0$

For roots to be equal
$\begin{matrix} b^{2} - 4 a c = 0 o r, b^{2} = 4 a c s o, (q)^{2} = 4 \times p \times r q^{2} = 4 p r \end{matrix}$

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Similar questions

Q. If the roots of

a x^{2} + b x + c = 0

p x^{2} + q x + r = 0

differ by the same quantity, then

\frac{b^{2} - 4 a c}{q^{2} - 4 p r}

α, β

are the roots of

a x^{2} + b x + c = 0

α + k, β + k

are the roots of

p x^{2} + q x + r = 0

\frac{b^{2} - 4 a c}{q^{2} - 4 p r}

x_{1}, x_{2}

be the roots of

a x^{2} + b x + c = 0

x_{3}, x_{4}

be the roots of

p x^{2} + q x + r = 0

x_{1}, x_{2}, \frac{1}{x_{3}}, \frac{1}{x_{4}}

are in A.P., then

\frac{b^{2} - 4 a c}{q^{2} - 4 p r}

x_{1}, x_{2}

be the roots of

a x^{2} + b x + c = 0

x_{3}, x_{4}

be the roots of

p x^{2} + q x + r = 0

.

x_{1}, x_{2}, \frac{1}{x_{3}}, \frac{1}{x_{4}}

are in A.P, then

\frac{b^{2} - 4 a c}{q^{2} - 4 p r} = ?

{p x}^{2} + q x + r = 0

has real roots, then

r = ?

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