Question

7. If
$\alpha and \beta$
are the zeros of the quadratic polynomial f(x) = x²– px + q then find the value of
(i)
${ \alpha }^{2} + { \beta }^{2}$

(ii)
$\binom{1}{ \alpha } + \binom{1}{ \beta }$
​

Answer 1

Answer:-

Given Polynomial: x² - px + q

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

We know that,

Sum of the zeroes = - b/a

$\sf \implies \: \alpha + \beta = \frac{ - ( - p)}{1} \\ \\ \implies \sf \large{\alpha + \beta = p \: \: - \: \: equation \: - \: (1)}$

And,

Product of the zeroes = c/a

$\sf \:\implies \large{ \alpha \beta = \frac{q}{1} \: \: - \: \: equation \: - \: (2)}$

We have to find:

$\sf 1) \: \large{\alpha ^ 2 + \beta ^ 2}$

We know that,

(a + b)² = a² + b² + 2ab

→ a² + b² = (a + b)² - 2ab

Hence,

$\sf \implies \alpha^2 + \beta^2 = {(\alpha + \beta)}^{2} - 2\alpha \beta$

Putting the values from equation (1) & (2) we get,

$\sf \implies \alpha ^ 2 + \beta^2 = {(p)}^{2} - 2(q) \\ \\ \sf \implies \large\red{{\alpha^2 + \beta^2 ={ p}^{2 }- 2q}}$

$\sf 2) \: \large{\frac{1}{\alpha} + \frac{1}{\beta}}$

Taking LCM we get,

$\sf \implies \frac{\beta + \alpha}{\alpha\beta}$

Putting the values we get,

$\sf \implies \large\red{ { \frac{p}{q} }}$

Answer 2

$\large\mathcal{\underbrace{\red{SOLUTION:-}}}$

GIVEN :-

• $\rm{\red{\alpha}\:and\:\red{\beta}}$ are the zero of the quadratic polynomial f(x) .

• where, f(x) = x² - px + q .

CONCEPT :-

✍️ If, two zeros of a quadratic polynomial are given, then

$\checkmark\:\rm{\boxed{Sum\:of\:the\:zeros\:=\:\dfrac{-(coefficient\:of\:x)}{coefficient\:of\:x^2}\:}}$

$\checkmark\:\rm{\boxed{Product\:of\:zeros\:=\:\dfrac{constant\:term}{coefficient\:of\:x^2}\:}}$

CALCULATION :-

✴️ Here,

• coefficient of ‘x²’ = 1

• coefficient of ‘x’ = -p

• constant term = q

$\longrightarrow\:\rm{\boxed{\alpha\:+\:\beta\:=\:\dfrac{-(-\:p)}{1}\:=\:p\:}}$

$\longrightarrow\:\rm{\boxed{\alpha\:\beta\:=\:\dfrac{q}{1}\:=\:q\:}}$

[1] $\rm{{(\alpha\:+\:\beta)}^2\:=\:{\alpha}^2\:+\:{\beta}^2\:+\:2\:\alpha\:\beta\:}$

$\rm{\implies\:{\alpha}^2\:+\:{\beta}^2\:=\:(\alpha\:+\:\beta)^2\:-\:2\:\alpha\:\beta\:}$

$\rm{\implies\:{\alpha}^2\:+\:{\beta}^2\:=\:(p)^2\:-\:2\times{q}\:}$

$\bigstar\:\rm{\purple{\boxed{\implies\:{\alpha}^2\:+\:{\beta}^2\:=\:p^2\:-\:2q\:}}}$

[2] $\rm{\dfrac{1}{\alpha}\:+\:\dfrac{1}{\beta}\:}$

$\rm{\implies\:\dfrac{\beta\:+\:\alpha}{\alpha\:\beta}\:}$

$\rm{\implies\:\dfrac{p}{q}\:}$

$\bigstar\:\rm{\blue{\boxed{\dfrac{1}{\alpha}\:+\:\dfrac{1}{\beta}\:=\:\dfrac{p}{q}\:}}}$