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Roots of a Quadratic Equation

If x-1 and ...

Question

If $(x - 1)$ and $(x - 2)$ are factors of $x^{4} - (p - 3) x^{3} - (3 p - 5) x^{2} + (2 p - 9) x + 6$ then the value of $p$ is:

1

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2

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3

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4

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Solution

The correct option is B $3$
The zero of any one of the factors will satisfy the equation:
$x^{4} - (p - 3) x^{3} - (3 p - 5) x^{2} + (2 p - 9) x + 6 = 0$
Select $(x - 1) = 0$ , which implies $x = 1$
Using this in $f (x) = x^{4} - (p - 3) x^{3} - (3 p - 5) x^{2} + (2 p - 9) x + 6 = 0$
we get $f (1) = 1^{4} - (p - 3) 1^{3} - (3 p - 5) 1^{2} + (2 p - 9) 1 + 6 = 0$
which implies, $p = 3$

Suggest Corrections

Similar questions

(x - 1), (x + 1)

(x - 2)

x^{4} + (p - 3) x^{3} - (3 p - 5) x^{2} + (2 p - 9) x + 6

p

(x + 1)

x^{4} + (p - 3) x^{3} - (3 p - 5) x^{2} + (2 p - 9) x + 6

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p

x + 1

x^{4} + (p - 3) x^{3} - (3 p - 5) x^{2} + (2 p - 9) x + 6

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P (x) = (\frac{6}{x}) p^{x} q^{6 - x}, x = 0, 1, 2, . . ., 6

2 P (2) = 3 P (3)

p =

∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 + p & 1 + p + q 2 & 3 + 2 p & 4 + 3 p + 2 q 3 & 6 + 3 p & 10 + 6 p + 3 q \end{matrix} ∣ ∣ ∣ ∣

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