Question

If a and B are the zeros of the polynomial f(x)= x² + px+q, form a polynomial whose zeros are (a+b)(a+b)and(a-b)(a-b)​

Answer 1

Answer:

let 2 zeroes be a and b of  polynomial  x² + px + q = 0

sum of roots = a + b = -p/1 = -p

products of roots = ab = q/1 = q

(a + b)² = a² + b² + 2ab ⇒  a² + b² = p² - 2q    

(a-b)² = a² + b² -2ab =  p² - 2q -2q = p² - 4q    

now ques asks for new quadratic eq whose roots are (a+b) ²  and  (a-b)²

so sum of new roots are = (a +b)² + (a-b)² = p² + p² -4q = 2p² - 4q

and  product of roots =  (a+b)²(a-b)² =  (p²)² (p²-4q)² = p⁴ (p⁴ +16q² + 8p²q)

hence new quadratic eq gonna be =

x² - x(sum of roots) + (products of roots)

x² - x(2p² - 4q) + p⁴(p⁴ + 16q² +8p²q) = 0

further  u can simplify.

hope my ans is correct ...........doubts , then enquire.

Answer 2

$k({x}^{2} - 2( {p}^{2} - 2q)x + {p}^{2}( {p}^{2} - 4 q))$

Step-by-step explanation:

$let \: \alpha \: and \: \beta \: be \: the \: zeroes \: of \: polynomial \: f(x) = {x}^{2} + px + q \\ \alpha + \beta = - p \\ \alpha \beta = q$

${ \alpha }^{2} + { \beta }^{2} = ( { \alpha + \beta })^{2} - 2 \alpha \beta \\ { \alpha }^{2} + { \beta }^{2} = ({p})^{2} - 2q \\ { \alpha }^{2} + { \beta }^{2} = {p}^{2} - 2q \\ sum = ( { \alpha + \beta })^{2} + ( { \alpha - \beta )}^{2}$

$= {p}^{2} - 2q + 2q + {p}^{2} - 2q - 2q \\ = 2 {p}^{2} - 4p \\ 2( {p}^{2} - 2q) \\ product \: = ( { \alpha + \beta })^{2} \times ( { \alpha - \beta })^{2} \\ = ({p} ) {}^{2} \times ( {p}^{2} - 2q - 2q) \\ = {p}^{2} ( {p}^{2} - 4q) \\ polynomial = k( {x}^{2} + (sum) x + product \\ = k( {x}^{2} - 2( {p}^{2} - 2q) x + {p}^{2} ( {p}^{2} - 4q)) \: where \: k \: is \: any \: non - zero \: constant$