If the roots of $a{x^2} + bx + c = 0\;{\rm{are}}\;\alpha ,\;\beta $ and the roots of $A{x^2} + Bx + C = 0\;{\rm{are}}\;\alpha - k,\;\beta - k,\;{\rm{then}}\;\frac{{{B^2} - 4AC}}{{{b^2} - 4ac}}$ is equal to

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Analyzing +4 | -1

If the roots of

a{x^2} + bx + c = 0\;\mathrm{are}\;\alpha ,\;\beta

and the roots of

A{x^2} + Bx + C = 0\;\mathrm{are}\;\alpha - k,\;\beta - k,\;\mathrm{then}\;\frac{{{B^2} - 4AC}}{{{b^2} - 4ac}}

is equal to

View answer and explanation

Correct option: C

{\left( {\frac{A}{a}} \right)^2}

Option C (

{\left( {\frac{A}{a}} \right)^2}

) is correct because the discriminant ratio is the square of the ratio of the leading coefficients, derived from the relationship between roots and coefficients.

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)