Solve x^2 + 5x + 6 = 0.

The first and the simplest approach to solve a quadratic equation is the splitting the middle term method. Not all quadratic equations can be solved by this method. When this method fails, the Quadratic Formula should be applied.

Here, we have

x^2 + 5x + 6 = 0
=> x^2 + 3x + 2x + 6 = 0
=> x(x + 3) + 2(x + 3) = 0
=> (x + 2)(x + 3) = 0
=> x = -2, x = -3

Hence, the solutions for x^2 + 5x + 6 = 0 are x = -2 and x = -3.

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

Roots of equation (x^2 - bx) / (ax - c) = (p - 1) / (p + 1) are equal in magnitude but opposite in sign. Find p.

The equation given is

(x^2 - bx) / (ax - c) = (p - 1) / (p + 1)

=> (p + 1)(x^2 - bx) = (p - 1)(ax - c)

=> (p + 1)x^2 - bpx -bx = apx - ax - pc + c

=> (p + 1)x^2 + ax - bx - apx - bpx + pc - c = 0

=> (p + 1)x^2 + (a - b - ap - bp)x  + pc - c = 0

It is given that the roots are equal in magnitude but opposite in sign. Therefore, sum of them will be equal to 0. (for eg, t and -t are equal in magnitude but opposite in sign, and hence there sum will be equal to 0)

We know that Sum of roots of the equation ax^2 + bx + c = 0 is -b/a. So, here,

Sum of roots = -(a - b - ap - bp)/(p + 1)

=> 0 = -(a - b - ap - bp)/(p + 1)

=> 0 =  -(a - b - ap - bp)

=> 0 = (a - b - ap - bp)

=> ap + bp = a - b

=> p(a + b) = (a - b)

=> p = (a - b)/(a + b)

The value of p will be (a - b)/(a + b) if the roots are equal in magnitude but opposite in sign.

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.

α, β are roots of x^2 + px + 1 = 0 and γ, δ are roots of equation x^2 + qx + 1 = 0, show that (α - γ)(β - γ)(α + δ)(β + δ) = q^2 - r^2

x^2 + px + 1 = 0 has roots α, β

x^2 + qx + 1 = 0 has roots γ, δ

Using the sum and product of roots formulae, we'll get

α + β = -p ; αβ = 1
γ + δ = -q ; γδ = 1

We have to prove that 

(α - γ)(β - γ)(α + δ)(β + δ) = q^2 - r^2

To begin with Left Hand Side,

LHS

= (α - γ)(β - γ)(α + δ)(β + δ)

This is a product of four expressions. Direct simplification can be a lot difficult, and can even be useless as it won't be easy to draw anything useful from it. We will begin with multiplying expressions in pairs of two.

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).

The ratio of roots of the equation x^2 + ax + a + 2 = 0 is 2. Find the value of a.

The equation given is x^2 + ax + a + 2 = 0 and it is told that the 'ratio' of its roots is 2. 

Now, if you think it over, you'll understand that if two numbers are in ratio 2, then one of those numbers have to be 2 times the other. Or, in other words we can say, that one of the number will be half (1/2 times) the other. Both these conditions are the same.

So, we can assume the roots of x^2 + ax + a + 2 = 0 to be t and 2t. 

(You can see it here. 2t is 2 times t, and obviously then, t is 1/2 times 2t)

So, here,

Sum of roots = -(a)/1

=> t + 2t = -a

=> 3t = -a

=> t = -a/3

Product of roots = (a + 2)/1

=> t * 2t = (a + 2)

=> 2t^2 = (a + 2)

If p, q are roots of the equation ax^2 + bx + c = 0, then find the equation whose roots are (2p + 3q) and (3p + 2q)

Given that p and q are roots of equation ax^2 + bx + c = 0

So, Sum of roots = -b/a
=> p + q = -b/a                  - (Equation 1)

Product of roots = c/a
=> pq = c/a                        - (Equation 2)

Now, we have to find the equation whose roots are (2α + 3) and (3α + 2).

Sum of roots of that equation will be
(2p + 3q) + (3p + 2q)
= (5p + 5q)
= 5(p + q)


Putting the value of (p + q) from Equation 1,
= 5(-b/a) = -5b/a
 
Product of roots of that equation will be

(2p + 3q) (3p + 2q)
= 6p^2 + 9pq + 4pq + 6q^2
= 6 (p^2 + q^2) + 13pq 
= 6 [(p + q)^2 - 2pq)] + 13pq
= 6(p + q)^2 - 12pq + 13pq
= 6(p + q)^2 + pq

Substituting the values of (p + q) and pq from Equation 1 and 2, 

= 6(-b/a)^2 + c/a
= 6(b^2/a^2) + c/a
= (6b^2 + ac) / a^2

We know that an equation whose sum of roots is m and product of roots is n can be written as

x^2 - mx + n = 0

So, the equation we want is

x^2 - (-5b/a) + [(6b^2 + ac) / a^2] = 0

Multiplying both the sides with a^2,

(a^2)x^2 + 5abx + 6b^2 + ac = 0 is the final answer.




    

What is the value of p for which the difference between the roots of the equation x^2 + px + 8 = 0 is 2?

We have the equation x² + px + 8 = 0. Given that difference of roots of this equation is 2.

We know that difference of roots = √(b²-4ac) / a or (√D)/a
So here,
=> Difference of roots = √(b²-4ac) / a
=> 2 = √(p²-4*1*8) / 1
=> 2 = √(p²-32)

Squaring both the sides,

=> 4 = p²-32  
=> p² = 36
=> p = +/- 6

So, the values of p for which difference between the roots of equation x² + px + 8 = 0 is 2 are 6 and -6.