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Standard X

Mathematics

Roots of a Quadratic Equation

If bi-quadrat...

Question

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 - 2p = 6, pq = - 21

P = - 3 q = 7

Which satisfies pq = - 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 = - 3 q = 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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