JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Application +4 | -1

(\alpha, \beta)

are roots of

\mathrm{ax}^2+2 \mathrm{bx}-\mathrm{a}=0

and quadratic equation whose roots are

\left(2 \alpha-\frac{1}{\beta}\right)

and

\left(2 \beta-\frac{1}{\alpha}\right)

\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0

, then

\mathrm{p}+\mathrm{q}+\mathrm{r}=

View answer and explanation

Correct option: C

6 \mathrm{~b}-8 \mathrm{a}

Option C (

6 \mathrm{~b}-8 \mathrm{a}

) is correct because it is the value of p+q+r obtained after finding the quadratic equation with the given roots and comparing coefficients.

More Questions from QUADRATIC EQUATIONS

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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,

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

{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\;

\;a \ne 0,

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

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