Question

if p&q are the rots of x^2+2x+1=0 then find the value of p^3+q^3​

Answer 1

f(x) = x² + 2x + 1 = 0

a = 1 ; b = 2 ; c = 1

If p & q are the roots of f(x),

then,

p + q = -b/a = -2/1 = -2

pq = c/a = 1/1 = 1

(p + q)² = (-2)²

=> p² + q² + 2pq = 4

=> p² + q² + 2 • 1 = 4

=> p² + q² = 4 - 2 = 2

=> p² + q² - pq = 2 - 1 = 1

Now,

p³ + q³ = (p + q)(p² + q² - pq)

p³ + q³ = -2 • 1

p³ + q³ = -2

Let's check.

Roots of f(x) = x² + 2x + 1 = 0 are equal, and is -1.

x² + 2x + 1 = (x + 1)²

Thus, p = q = -1.

=> p³ + q³ = (-1)³ + (-1)³ = - 1 - 1 = -2

Hence checked!!!