JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

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Analyzing +4 | -1

\alpha

and

\beta

are the roots of

{x^2} + px + 1 = 0

and

\gamma

and

\delta

are the roots of

{x^2} + qx + 1 = 0,

then the value of

\left( {\alpha - \gamma } \right)\left( {\beta - \gamma } \right)\left( {\alpha + \delta } \right)\left( {\beta + \delta } \right),

View answer and explanation

Correct option: B

{q^2} - {p^2}

Option B (

{q^2} - {p^2}

) is correct because by substituting the roots into the given expression and using the relationships between the roots and coefficients of the quadratic equations, the expression simplifies to the difference of squares of q and p.

More Questions from QUADRATIC EQUATIONS

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The solution of the equation

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

is If

\alpha ,\beta

are the roots of

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

and

\alpha < 1 < \beta ,

then If

\alpha ,\;\beta

be the roots of the equation

{x^2} - px + q = 0

and

{\alpha _1},{\beta _1}

be the roots of the equation

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

then the equation whose roots are

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

is If

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

for all values of

x

, then

a

lies in the interval If

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

and

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

and

\;a \ne 0,

then Root(s) of the equation

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

belonging to the domain of definition of the function

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

is (are)