If $a$ and $b$ are roots of ${x^2} - px + q = 0$, then $\frac{1}{a} + \frac{1}{b} = ?$

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

a

and

b

are roots of

{x^2} - px + q = 0

, then

\frac{1}{a} + \frac{1}{b} = ?

View answer and explanation

Correct option: D

\frac{p}{q}

Option D (

\frac{p}{q}

) is correct because it is derived by expressing the sum of the reciprocals of the roots in terms of the sum and product of the roots, which are related to the coefficients of the quadratic equation.

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)