JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

If the equations

x^{2}+a x+b=0

and

x^{2}+a^{\prime} x+b^{\prime}=0

have a common root, then this common root is equal to

View answer and explanation

Correct option: D

\frac{\left(b-b^{\prime}\right)}{\left(a^{\prime}-a\right)}

Option D (

\frac{{\left(b-b^{\prime}\right)}}{{\left(a^{\prime}-a\right)}}

) is correct because by subtracting the two equations, the common root can be isolated and solved for.

More Questions from QUADRATIC EQUATIONS

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

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\alpha < 1 < \beta ,

\alpha ,\;\beta

{x^2} - px + q = 0

{\alpha _1},{\beta _1}

{x^2} - qx + p = 0,

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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)