If the difference between the roots of ${x^2} + ax + b = 0\,{\rm{and}}\,{x^2} + bx + a = 0$ is same and $a \ne b,$ then

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

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Analyzing +4 | -1

If the difference between the roots of

{x^2} + ax + b = 0\,\mathrm{and}\,{x^2} + bx + a = 0

is same and

a \ne b,

then

View answer and explanation

Correct option: A

a + b + 4 = 0

Option A (

a + b + 4 = 0

) is correct because the condition on the difference of roots leads to this relationship between a and b.

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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)