Math problem?

Question

All 2s should be superscripted


The roots of the equation x2 - 3mx + m2 = 0 are p and q where p > q. Find the values of p2 + q2 and p - q in terms of m when m is positive.


I managed to find the value of p2 + q2 i.e. 7m2.
Can you show me how to solve the second value

Answer 1

So
x² + 3mx + m² = (x-p)(x-q) = x² + (-p - q)x + pq.

Setting the coefficients equal on both sides, we get

3m = -p - q (x coefficient)
m² = pq (constant term)

So p² + q²
= (p+q)² - 2pq
= (-p - q)² - 2pq
= (3m)² - 2m²
= 7m² as you said.

Now for the second part, we'll go for (p-q)² instead, and take the square root of this to get at p-q.
(p-q)²
= p² + q² - 2pq
= 7m² - 2m²
= 5m²

So p - q = (sqrt5)m (we take the positive root because p>q).

Answer 2

it;s simple see
since p and q are roots of d equation, so
sum of the roots = -(middle term/first term)
here middle term is -3m.right? and first term is 1 and also
product of the roots = cofficent of last term/ cofficent of first term
so sum of roots = p+q = 3m/1=3m
on squaring both sides , we get
(p+q)^2 =(3m)^2
p^2 + q^2 + 2pq = 9m^2
product of roots = p x q = m^2
put d value of p x q in above equation...we get
p^2 + q^2 + 2(m^2) = 9m^2
so p2 + q2 = 7m^2....answer

Answer 3

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Answer 4

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