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.

r is a root of equation x^2 - 3x + 4 = 0. Reduce (r^3 - 4r^2 + 9r - 2) into a linear expression without actually calculating the value of r.

Given that r is a root of equation x^2 - 3x + 4 = 0, which means r should satisfy the equation. So we can say that

(r)^2 - 3(r) + 4 = 0
=> r^2 - 3r + 4 = 0               (1)

Multiplying r on both sides of the equation, we'll get

=> r (r^2 - 3r + 4) = 0 * r
=> r^3 - 3r^2 + 4r = 0          (2)

Now, using (1) and (2), we'll try reducing r^3 - 4r^2 + 9r - 2

=> r^3 - 4r^2 + 9r - 2
=> (r^3 - 3r^2 + 4r) - r^2 + 5r - 2
=> 0 - r^2 + 5r - 2          (by using 2)
=> - (r^2 - 5r + 2)
=> - (r^2 - 3r + 4 - 2r + 2)
=> - (0 - 2r + 2)              (by using 1)
=> 2r - 2
=> 2(r - 1)

which is linear.

x^2 + px - q = 0 has roots α and β, x^2 + px + r = 0 has roots γ, δ. Prove that (α - γ)(α - δ) = (β - γ)(β - δ) = q + r

The information given is, 

x^2 + px - q = 0 has roots α, β
x^2 + px + r = 0 has roots γ, δ

Using Sum of roots and Product of roots formulae on both these equations, we get these results-

α + β = -p               (1)
αβ = -q                   (2)

γ + δ = -p               (3)
γδ = r                     (4)


Lets have a look at what we have to prove.

(α - γ)(α - δ) = (β - γ)(β - δ) = q + r

We will begin with (α - γ)(α - δ).

px^2 + qx + r = 0 has roots sin(theta) and cos(theta). Prove that p^2 - q^2 + 2pr = 0

The equation given is px^2 + qx + r = 0. Its roots are sinθ and cosθ.

For this equation,

Product of roots = r/p

=> sinθ.cosθ = r/p

Sum of roots = -q/p

=> sinθ + cosθ = -q/p           (because the roots are sinθ and cosθ)

Squaring both the sides, we'll get

=> (sinθ + cosθ)^2 = (-q/p)^2

=> sin^2 θ + cos^2 θ + 2sinθ.cosθ = q^2/p^2

=> 1 + 2sinθ.cosθ = q^2/p^2        (using sin^2 θ + cos^2 θ = 1)

Now, here, we will substitute the value of sinθ.cosθ we calculated in the beginning.

=> 1 + 2(r/p) = q^2/p^2

=> (p + 2r)/p = q^2/p^2

=> (p + 2r) = q^2/p

=> p^2 + 2pr = q^2

=> p^2 - q^2 + 2pr = 0

Hence Proved.

If p, q ,r be the roots of the equation x(1 + x^2) + x^2(6 + x) + 2 = 0 then find the value of 1/p + 1/q + 1/r

We have the equation x(1 + x^2) + x^2(6 + x) + 2 = 0

=> x + x^3 + 6x^2 + x^3 + 2 = 0

=> 2x^3 + 6x^2 + x + 2 = 0

This is a cubic equation. A cubic equation is generally denoted by the form ax^3 + bx^2 + cx + d = 0

For a cubic Equation ax^3 + bx^2 + cx + d = 0, having roots j, k ,l,  we have the following general results:-

j + k + l = -b/a

jk + kl + lj = c/a

jkl = -d/a

Returning to our problem, we have 2x^3 + 6x^2 + x + 2 = 0 which has roots p, q, r. so

p + q + r = -6/2 = -3

pq + qr + rp = 1/2

pqr = -2/2 = -1

We have to evaluate 1/p + 1/q + 1/r

= (qr + pr + pq) / pqr

= (1/2) / (-1)

= -1/2

Hence, the value of 1/p + 1/q + 1/r is -1/2