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Discriminant

Discriminant ...

Question

Discriminant of a quadratic equation $p x^{2} + q x + r = 0$ is given by.

A

(q^{2} - 4 p r)

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B

\sqrt{q^{2} + 4 q r}

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C

\sqrt{q^{2} - 4 p r}

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D

q^{2} + 4 q^{2}

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Solution

The correct option is B $(q^{2} - 4 p r)$

Given equation is $p x^{2} + q x + r = 0$
Here $a = p$ , $b = q$ and $c = r$
Discriminant $D = b^{2} - 4 a c$ $= q^{2} - 4 p r$
Hence, option A is correct.

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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 equal roots, then

q^{2} =

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