(D) Let $f(x) = x^2+px-q^2$. Since the coefficient of $x^2$ is $1 > 0$,the parabola opens upwards. Given that one root $\alpha < 2$ and the other root

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(D) Let $f(x) = x^2+px-q^2$. Since the coefficient of $x^2$ is $1 > 0$,the parabola opens upwards. Given that one root $\alpha 2$,the value of the function at $x=2$ must be negative,i.e.,$f(2) Substituting $x=2$ into the quadratic expression: $f(2) = (2)^2 + p(2) - q^2 $4 + 2p - q^2 Thus,the correct option is $D$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Quadratic expressions and Position of roots

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If one root of $x^2+px-q^2=0$,where $p$ and $q$ are real,is less than $2$ and the other is greater than $2$,then:

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A
$4+2p+q^2 > 0$
B
$4+2p+q^2 < 0$
C
$4+2p-q^2 > 0$
D
$4+2p-q^2 < 0$

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4-2.Quadratic Equations and Inequations

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Quadratic expressions and Position of roots

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Match the following quadratic expressions with their minimum values:
Quadratic expression The minimum value
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