If the quadratic equation $x^{2} - 7 x - 98 = 0$ can be written as $(x + a) (x - 2 a) = 0,$
the value of $‘ a^{'}$ is equal to:

[1 Mark]

-7

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-14

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Solution

The correct option is B 7
Given,
$x^{2} - 7 x - 98 = (x + a) (x - 2 a)$ $= x^{2} - a x - 2 a^{2}$
By comparing the coefficients on both sides, we get,
$\Rightarrow - a = - 7 and - 2 a^{2} = - 98$
$\Rightarrow a = 7 and a^{2} = 49$
$\Rightarrow a = 7 and a = \pm 7$
$∴ a = 7$

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If the quadratic equation x2−7x−98=0 can be written as (x+a)(x−2a)=0, the value of ‘a′ is equal to: [1 Mark]

The correct option is B 7Given, x2−7x−98=(x+a)(x−2a) =x2−ax−2a2 By comparing the coefficients on both sides, we get, ⇒−a=−7 and −2a2=−98 ⇒a=7 and a2=49 ⇒a=7 and a=±7 ∴a=7

If the quadratic equation $x^{2} - 7 x - 98 = 0$ can be written as $(x + a) (x - 2 a) = 0,$
the value of $‘ a^{'}$ is equal to:

[1 Mark]

The correct option is B 7
Given,
$x^{2} - 7 x - 98 = (x + a) (x - 2 a)$ $= x^{2} - a x - 2 a^{2}$
By comparing the coefficients on both sides, we get,
$\Rightarrow - a = - 7 and - 2 a^{2} = - 98$
$\Rightarrow a = 7 and a^{2} = 49$
$\Rightarrow a = 7 and a = \pm 7$
$∴ a = 7$