JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

Let

\alpha

and

\beta

be the roots of

p x^2+q x+r=0

p \neq 0

. If

p, q, r

are in A.P. and

\frac{1}{\alpha}+\frac{1}{\beta}=4

, then value of

|\alpha-\beta|

View answer and explanation

Correct option: D

\frac{2}{9} \sqrt{13}

Option D (

\frac{2}{9} \sqrt{13}

) is correct because using the given conditions that p, q, r are in A.P. and the relationship between the roots, we can find the value of |

\alpha - \beta

More Questions from QUADRATIC EQUATIONS

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

\alpha ,\beta

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\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

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