If bi-quadratic equation $x^{4} - 2 x^{3} + 4 x^{2} + 6 x - 21 = 0$ can also be written as $(x^{2} + p) (x^{2} - 2 x + q) = 0$ . Find the sum of $p + q$ .

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Solution

$(x^{2} + p) (x^{2} - 2 x + q) = x^{4} - 2 x^{3} + (p + q) x^{2} - 2 p x - p q = 0$ . . . $(1)$

Given equation,

$x^{4} - 2 x^{3} + 4 x^{2} + 6 x - 21 = 0$ . . . $(2)$

Compare the coefficients of $x^{4}, x^{3}, x^{2}, x$ and constant in both the equations

We get, $p + q = 4$
$- 2 p = 6$ ,
$p q = - 21$

$p = - 3 q = 7$

Which satisfies $p q = - 21$

$p + q = 4$

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If bi-quadratic equation x4−2x3+4x2+6x−21=0 can also be written as (x2+p)(x2−2x+q)=0. Find the sum of p+q. __

(x2+p)(x2−2x+q)=x4−2x3+(p+q)x2−2px−pq=0 . . . (1) Given equation, x4−2x3+4x2+6x−21=0 . . . (2) Compare the coefficients of x4,x3,x2,x and constant in both the equations We get, p+q=4 −2p=6, pq=−21 p=−3q=7 Which satisfies pq=−21 p+q=4

If bi-quadratic equation $x^{4} - 2 x^{3} + 4 x^{2} + 6 x - 21 = 0$ can also be written as $(x^{2} + p) (x^{2} - 2 x + q) = 0$ . Find the sum of $p + q$ .

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