JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Analyzing +4 | -1

a,b

are odd integers, then the roots of the equation

2\;a{x^2} + \left( {2\;a + b} \right)x + b = 0,a \ne 0

are

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Correct option: A

Rational

Option A is correct because the roots of the given quadratic equation are rational, as they can be expressed as -1 and -b/2a.

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

\alpha ,\beta

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

\alpha < 1 < \beta ,

\alpha ,\;\beta

{x^2} - px + q = 0

{\alpha _1},{\beta _1}

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

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

x

a

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

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

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

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