Question

# Maths Time!

Question :-

Find quadratic equation such that its roots are square of sum of the roots and square of difference of the roots of equation:
2x² + 2(p + q)x + p² + q² = 0

All the best! :)
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Answer 1

$\huge{\boxed{\boxed{\mathfrak{\red{An swer:-}}}}}$

${\bold{\underline{\underline{To \: find:-}}}}$

•A quadratic equation = ❓

✴let the roots of this equation be M and N.

${\bold{\underline{\underline{In \: given\: equation :-}}}}$

✴Let the roots of this given equation be A and B.

✴Equation:-

$\implies\ 2{x}^{2}+2(p+q)x+{p}^{2}+{q}$

Now,

•Sum of roots ➡

$\implies\ A+B= {-2(p+q)}/{2}$

•Product of roots➡

$\implies\ AB ={p^2+q^2}/{2}$

____________♥

we have to find the:-

$\to\ Squares\: of \: sum \: of \: roots \\ \\ \to\ Squares \: of \: difference \: of \: roots$

Now;

➡$M= (a+b)^2= (p+q)^2$

➡$N=(a-b)^2=(-p-q)^2$

➡$M + N = 4pq$

➡$MN = (p+q)^2{-(p-q)^2 }=(-p^2-q^2)^2$

Required equation ➡

♥$\implies\ x^2-4(pq)x-[p^2-q^2]^2$♥

$\mathbb\pink{Thanks....}$

Answer 2

Let the roots of the required quation be M and N

let the roots of the equation 2x²+2(p+q)x+p²+q²=0 be a and b

a + b = -(p+q)

ab = (p^2 + q^2) / 2

(a+b)^2 = (p+q)^2

(a-b)^2 = (a+b)^2 - 4ab

(a-b)^2 = -(p - q)^2

we wanted the values of square of sum of the roots and square of difference of the roots

Now M = (a+b)^2 = (p+q)^2 and

N = (a-b)^2 = -(p - q)^2

M + N = 4pq

MN = (p+q)^2 [-(p - q)^2]

MN= -(p^2 - q^2)^2

hence the required equation is

x^2 - (4pq)x - (p^2 - q^2)^2 = 0

Hope this helps!!