Question

If p and q are the roots of the equation x²+2x+1=0 the the value of p³+q³ become​

Answer 1

Answer:

-2

Step-by-step explanation:

Given

$f(x) => x^2+2x+1 = 0$. It is in the form $ax^2+bx+c = 0$

given $p$ and $q$ are roots of $f(x)$

Sum of roots = $p+q =$ $-\frac{b}{a} = -\frac{2}{1} = -2 - (i)$

Product of roots = $pq=$ $\frac{c}{a} = \frac{1}{1} = 1 -(ii)$

We know,

$(a+b)^2 = a^2+b^2+2ab$

Substituting a and b with p and q

$(p+q)^2= p^2+q^2+2pq$

$(-2)^2 = p^2+q^2+2(1)\\4 = p^2+q^2 +2\\p^2+q^2 = 4-2\\p^2+q^2 = 2 - (iii)$

$p^3+q^3=?$

We know,

$a^3+b^3 = (a+b)(a^2+b^2-ab)$

Substituting, a and b with p and q

$p^3+q^3$$= (p+q)(p^2+q^2-pq)$

$=(-2)(2-1) [From (i),(ii),(iii), substituting]\\=(-2)(1) = -2$

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Answer 2

p and q are roots of x²+2x+1=0

f(x) = x² + 2x + 1 = 0

f(p) = p² + 2p + 1² = 0

= (p+1)² = 0

= p = -1

f(q) = q² + 2q + 1² = 0

= (q + 1)² = 0

= q = -1

p³ + q³

= (-1)³ + (-1)³

= (-1) + (-1)

= -2

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