Question

if p and q are the zeroes of the polynomial x^2+ax+b then show p^2+q^2=a^2-2b​

Answer 1

Step-by-step explanation:

p + q = -a/1. ........(1)

putting x= p

p^2 + a.p + b

p^2 = -a.p - b. .......(2)

putting x = q

q^2 + a.q + b

q^2 = -a.q - b. .......(3)

adding 2 & 3 equations

p^2 + q^2 = -a.p - a.q - b - b

p^2 + q^2 = -a(p+q) -2b

from equation 1

p^2 + q^2 = -a×-a -2b

p^2 + q^2 = a^2 -2b

Answer 2

Solution :-

$\implies \sf \: p + q = \frac{ - a}{1} ...(1)$

$\fbox{ \sf \: putting \: x = p}$

$\implies \sf \: p² + a.p + b$

$\implies \sf \: p² = a.p - b...(2)$

$\fbox{\sf \: putting \: x = q}$

$\implies \sf \: q² + a.q + b$

$\implies \sf \: q² = a.q - b...(3)$

adding 2 and 3 equation s

• p² + q² = -a.p - a.q - b - b
• p² + q² = -a( p + q ) - 2b

From eqaution one :-

• p² + q² = -a × -a - 2b
• p² + q² = a² - 2b

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★This question is exerted from chapter polynomials.

★ In this chapter always careful about the signs.

★ Do your best and be Brainly :)

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