JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Application +4 | -1

\alpha, \beta, \gamma

are the roots of the equation

\mathrm{x}^3+\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0

, then the cubic equation whose roots are

\alpha(\beta+\gamma), \beta(\gamma+\alpha), \gamma(\alpha+\beta)

is :

View answer and explanation

Correct option: A

\mathrm{x}^3-2 \mathrm{qx}^2+\left(\mathrm{q}^2+\mathrm{pr}\right) \mathrm{x}+\mathrm{r}^2-\mathrm{pqr}=0

Option A (

x^3-2 q x^2+\left(q^2+p r\right) x+r^2-p q r=0

) is correct because by substituting y = alpha(beta + gamma) and using Vieta's formulas, we can derive the required cubic equation.

More Questions from QUADRATIC EQUATIONS

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

\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,

\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)