Question

if x-a is a factor of x^2+ px-q and x^2+rx -t .then prove that a= t-q/r-p​

Answer 1

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$\underline{\large\mathcal\red{solution}}$

(x-a )is a factor of( x²+ px-q )and (x²+rx -t )

so the remainder in both cases when dividing by (x-a) is same i.e, 0

now ....

$a {}^{2} + p(a) - q = a {}^{2} + r(a) - t \\ = > pa - ra = q - t \\ = > a(p - r) = q - t \\ = > a = \frac{(q - t)}{(p - r)} \\ = > a = \frac{(t - q)}{(r - p)} \: \: (proved)$

$\large\mathcal\red{hope\: this \: helps \:you......}$

Answer 2

Step-by-step explanation:

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