If the roots of the equation $a{x^2} + bx + c = 0$ of the form $\frac{{k + 1}}{k}\;\,{\rm{and}}\,\;\frac{{k + 2}}{{k + 1}},\;\,{\rm{then}}\,\;{\left( {a + b + c} \right)^2}$ is equal to

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Analyzing +4 | -1

If the roots of the equation

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

of the form

\frac{{k + 1}}{k}\;\,\mathrm{and}\,\;\frac{{k + 2}}{{k + 1}},\;\,\mathrm{then}\,\;{\left( {a + b + c} \right)^2}

is equal to

View answer and explanation

Correct option: C

{b^2} - 4ac

The correct answer is C (

{b^2} - 4ac

) because by using the given roots and Vieta's formulas, we can derive the relationship and simplify to find the value of

(a+b+c)^2

Practice All Chapter MCQs →

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{\alpha _1},{\beta _1}

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{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)