JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Analyzing +4 | -1

Let

\alpha, \beta

be the roots of the equation

x^2-\sqrt{2} x+\sqrt{6}=0

and

\frac{1}{\alpha^2}+1, \frac{1}{\beta^2}+1, \frac{1}{\beta^2}+1

be the roots of the equation

x^2+a x+b=0

. Then the roots of the equation

x^2-(a+b-2) x+(a+b+2)

=0

are

View answer and explanation

Correct option: B

real and both negative

Option B (real and both negative) is correct because the roots of the quadratic equation

6x^2 + 17x + 7 = 0

are real and both negative.

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)