JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

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Analyzing +4 | -1

Let

a, b, c, p, q

be real numbers. Suppose

\alpha, \beta

are roots of the equation

x^2+2 p x+q=0

and

\alpha, \frac{1}{\beta}

are the roots of the equation

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

, where

\beta^2 \notin\{-1,0,1\}

.

Statement-1

\left(p^2-q\right)\left(b^2-a c\right) \geq 0

and

Statement-2

b \neq p a

c \neq q a

View answer and explanation

Correct option: B

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

Option B (Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1) is correct because both statements are individually true, but Statement-2 does not explain Statement-1.

More Questions from QUADRATIC EQUATIONS

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The solution of the equation

\log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0

is If

\alpha ,\beta

are the roots of

{x^2} - 3x + a = 0,a \in R

and

\alpha < 1 < \beta ,

then If

\alpha ,\;\beta

be the roots of the equation

{x^2} - px + q = 0

and

{\alpha _1},{\beta _1}

be the roots of the equation

{x^2} - qx + p = 0,

then the equation whose roots are

\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\mathrm{and}\;\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}},

is If

{x^2} + 4ax + 2 > 0

for all values of

x

, then

a

lies in the interval If

f\left( x \right) = a{x^2} + bx + c\;

and

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

and

\;a \ne 0,

then Root(s) of the equation

9{x^2} - 18\left| x \right| + 5 = 0

belonging to the domain of definition of the function

f\left( x \right) = \log \left( {{x^2} - x - 2} \right)

is (are)