JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

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Analyzing +4 | -1

Let

p, q

and

r

be real numbers

(\mathrm{p} \neq \mathrm{q}, \mathrm{r} \neq 0)

, such that the roots of the equation

\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}

are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to.

View answer and explanation

Correct option: B

p^{2}+q^{2}

Option B (

p^{2}+q^{2}

) is correct because by setting the sum of the roots to zero (since they are equal in magnitude but opposite in sign) and using the quadratic formula, we can find the sum of the squares of the roots.

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)