JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Analyzing +4 | -1

In writing an equation of the form

a{x^2} + bx + c = 0;

the coefficient of

x

is written incorrectly and roots are found to be equal. Again in writing the same equation the constant term is written incorrectly and it is found that one root is equal to those of the previous wrong equation while the other is double of it. If

\alpha

and

\beta

be the roots of correct equation, then

{\left( {\alpha - \beta } \right)^2}

is equal to

View answer and explanation

Correct option: B

5\,\alpha \;\beta

Option B (

5\\,\alpha \;\beta

) is correct because by analyzing the incorrect equations and relating their roots to the correct equation's roots, we can derive the relationship

{\left( {\alpha - \beta } \right)^2} = 5\alpha \beta

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)