Question

If m and n are the zeroes of the quadratic polynomials
F(x)=x^2-px+q, then find the values of : a) m^2+n^2 b) m^-1 + n^-1​

Answer 1

Answer:-

Given Polynomial => x² - px + q = 0

Let , a = 1 ; b = - p ; c = q

We know that,

Sum of the roots = - b/a

→ m + n = - (- p)/1 = p

And,

Product of the roots = c/a

→ mn = q/1 = q

a) m² + n² can be written as (m + n)² - 2mn

i.e., m² + n² + 2mn - 2mn.

→ m² + n² = (p)² - 2(q)

→ m² + n² = p² - 2q

b) m^(- 1) = 1/m and n^(- 1) = 1/n

Taking LCM,

1/m + 1/n = (n + m)/mn

→ (m + n)/mn

Putting values,

→ p/q

→ m^(- 1) + n^(- 1) = p/q

Answer 2

If $\alpha$ and $\beta$ are roots of the quadratic polynomial $p(x)=ax^2+bx+c,$ then,

• $\alpha+\beta=-\dfrac{b}{a}$

• $\alpha\beta=\dfrac{c}{a}$

Since $m$ and $n$ are roots of the quadratic polynomial $F(x)=x^2-px+q,$ we have,

$\longrightarrow m+n=p$

$\longrightarrow mn=q$

Then,

$\longrightarrow m^2+n^2=m^2+n^2+2mn-2mn$

$\longrightarrow m^2+n^2=(m+n)^2-2mn$

$\longrightarrow\underline{\underline{m^2+n^2=p^2-2q}}$

And,

$\longrightarrow m^{-1}+n^{-1}=\dfrac{1}{m}+\dfrac{1}{n}$

$\longrightarrow m^{-1}+n^{-1}=\dfrac{m+n}{mn}$

$\longrightarrow\underline{\underline{m^{-1}+n^{-1}=\dfrac{p}{q}}}$