(a) (4m)x^2 - 2(m + n)x + n = 0 (b) (a + b + c)x^2 - 2(a + b) + (a + b - c) = 0 -- Prove that these equations have rational roots.

For roots of any quadratic equation to be rational, discriminant D must be a perfect square.

We know that for quadratic equation ax^2 + bx + c = 0, D = b^2 - 4ac

(a) For (4m)x^2 - 2(m + n)x + n = 0

D

= [-2(m + n)]^2 - 4(4m)(n)

= 4(m + n)^2 - 16mn

= 4(m^2 + n^2 + 2mn) - 16mn

= 4m^2 + 4n^2 + 8mn - 16mn

= 4m^2 + 4n^2 - 8mn

= 4(m^2 + n^2 - 2mn)

= 4(m - n)^2

= 2^2 . (m - n)^2

= (2m - 2n)^2 

which is a perfect square. Since discriminant of this equation is a perfect square, it will have rational roots.

Roots of equation x^2 - rx + m = 0 differ by 1. Prove that r^2 = 4m + 1

The equation is x^2 - rx + m = 0 and the difference between its roots is 1.

We know that for the equation ax^2 + bx + c = 0, 

Difference of roots = √D /a 

where D is the Discriminant of the equation, which equals (b^2 - 4ac). So, here,

Difference of roots = √[(-r)^2 - 4(1)(m)] / 1

=> 1 = √(r^2 - 4m)

Squaring both the sides,

=> 1 = r^2 - 4m

=> r^2 = 4m + 1

Hence Proved

If the ratio of roots of the equation x^2 + px + q = 0 is equal to the ratio of roots of x^2 + rx + m = 0, prove that m.p^2 = q.r^2

First of all, lets have a look at what we're given. Ratio of roots of x^2 + px + q = 0 and x^2 + rx + m = 0 are equal.

We are assuming that the roots of x^2 + px + q = 0 are α, β while roots of x^2 + rx + m = 0 are γ, δ. So, as we're given,

α / β = γ / δ

By using the sum and product of roots formulae, we can say that

α + β = -p ; αβ = q
γ + δ = -r ; γδ = m

We have to prove that m.p^2 = q.r^2. Now, there can be several approaches to the proof. We'll be telling you two of them.

For equation ax^2 + bx + c = 0, find the condition such that one root may be (i) 4 times the other (ii) m times the other (iii) more than the other by k

The equation given is ax^2 + bx + c = 0

(i) One root is 4 times the other. We can assume them as t and 4t.

Sum of roots = -b/a
=> t + 4t = -b/a
=> 5t = -b/a
=> t = -b/5a               (equation 1)

Product of roots = c/a
=> t * 4t = c/a
=> 4t^2 = c/a

Substituting the value of t from equation 1, we'll get

=> 4(-b/5a)^2 = c/a
=> (4b^2)/(25a^2) = c/a
=> (4b^2)/25a = c
=> 4b^2 = 25ac

Find the set of possible values of 'm' for which x^2 - (m^2 - 5m + 5)x + (2m^2 - 3m - 4) = 0 has roots whose sum and product are both less than 1.

We have the equation x^2 - (m^2 - 5m + 5)x + (2m^2 - 3m - 4) = 0;

Sum of roots = (m^2 - 5m + 5)
Product of roots = (2m^2 - 3m - 4)

Given that both the sum and product of roots are < 1.

So,

m^2 - 5m + 5 < 1
=> m^2 - 5m + 4 < 0
=> m^2 - m - 4m + 4 < 0
=> m(m - 1) - 4(m - 1) < 0
=> (m - 4)(m - 1) < 0

To satisfy this, m should lie in (1, 4)

Also,

2m^2 - 3m - 4 < 1
=> 2m^2 - 3m - 5 < 0
=> 2m^2 + 2m - 5m - 5 < 0
=> 2m(m + 1) - 5(m + 1) < 0
=> (m + 1)(2m - 5) < 0

To satisfy this, m should lie in (-1, 5/2)

To satisfy both, our answer should be the intersection of (1, 4) and (-1, 5/2) which comes out to be (1, 5/2)