JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

a+b \neq 0

and the roots of

x^2-p x+q=0

differ by -1 , then

p^2-4 q

equals:

View answer and explanation

Correct option: C

Option C (1) is correct because the difference of roots being -1 allows us to relate the sum and product of roots, leading to the value of p² - 4q.

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\alpha ,\beta

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\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)