If alpha and beta are the roots of equation x^2 = x + 1, then find the value of alpha^2/beta - beta^2/alpha.

We are given the equation

x^2 = x + 1
=> x^2 - x - 1 = 0

As it is given that the roots of this equation are α, β, we can, by the use of the sum and product of roots formulae, say that,

α + β = -(-1)/1 = 1

αβ = -1

Now, we are asked to find the value of

 α^2/β - β^2/α.

= (α^3 - β^3) / αβ

= (α - β)(α^2 + αβ + β^2) / -1

= -√[(α + β)^2 - 4αβ] [(α + β)^2 - αβ]

= -√[1 - 4(-1)] (1 - (-1))

= -2√5

If a and b are the root of 10x^2 - 13x +3 =0 find a+b (2) ab (3) a^2+b^2 (4)a^3+b^3 (5) a-b

a and b are the roots of 10x^2 - 13x + 3 = 0. By using the sum and product of roots formulae, we'll get

a + b = -(-13)/10 = 13/10
ab = 3/10

This itself solves (1) and (2).

(3) a^2 + b^2
= (a + b)^2 - 2ab
= (13/10)^2 - 2(3)/10
= 169/100 - 6/10
= 109/100

(4) a^3 + b^3
= (a + b)(a^2 - ab + b^2)
= (a + b) ((a + b)^2 - 3ab)
= 13/10 ((13/10)^2 - 3(3/10))
= 13/10 (169/100 - 9/10)
= 13/10 (79/100)
= 1027/1000

(5) a - b
= √(a - b)^2
= √(a^2 + b^2 - 2ab)
= √[(a + b)^2 - 4ab]
= √[(13/10)^2 - 4(3/10)]
= √(169/100 - 12/10)
= √49/100
= 7/10

If p and q are the roots of the equation x^2 - px + q = 0, find the value of p and q.

It is given that the equation x^2 - px + q = 0 has roots p and q. We can begin solving either by using the sum and product formulae, or by simply substituting the roots into the equation and then work with the resultant equations. Either way, the approach is more or less the same because the Sum and Product of roots formulae are nothing but derivations made by substituting roots into the equation for the standard equation.

Here, 

Sum of roots = p
=> p + q = p
=> q = 0

Product of roots = q
=> pq = q
=> pq - q = 0
=> q (p - 1) = 0
=> q = 0, p = 1

The values of p and q hence found are p = 1, and q = 0. This makes the original equation x^2  - x = 0.

If alpha and beta are roots of ay^2 + by + c = 0, Find the value of 1/alpha^4 + 1/beta^4

ay^2 + by + c = 0 has roots α, β. For this equation, we know that

Sum of roots = -b/a
=> α + β = -b/a

Product of Roots = c/a
= αβ = c/a

Using these two results, we'll find out the value of 1/α^4 + 1/β^4

= 1/α^4 + 1/β^4
= (α^4 + β^4) / α^4.β^4
= (α²)² + (β²)² / (αβ)^4
= (α² + β²)² - 2α²β² / (αβ)^4               by using a² + b² = (a + b)² - 2ab in the numerator
= ((α + β)²  - 2αβ)² - 2(αβ)² / (αβ)^4               by using the same identity again

We will now substitute the values of α + β and αβ which we found earlier

Equation (a - b)x^2 - (b + c - a)x - c = 0, find its nature of roots.

The equation given is (a - b)x^2 - (b + c - a)x - c = 0. To determine the nature of roots, we'll first find the discriminant of the equation. We know that for ax^2 + bx + c = 0, discriminant D = b^2 - 4ac

So, here,

D

= [-(b + c - a)]^2 - 4(a - b)(-c)

= b^2 + c^2 + a^2 + 2bc - 2ac - 2ab + 4ac - 4bc

= b^2 + c^2 + a^2 - 2bc + 2ac - 2ab

If you look at this expression carefully, it is in form of a perfect square.

= (-b)^2 + (c)^2 + (a)^2 + 2(-b)(c) + 2(c)(a) + 2(a)(-b)

= (-b + c + a)^2

So, D = (-b + c + a)^2 for the given equation. Lets make our conclusions now.

Since D is a perfect square, the equation will have rational roots. 

Again, since (-b + c + a)^2 is a perfect square, it is always greater than or equal to 0. The roots will be real. (of course, rational numbers are always real)

Show that equation x^2 + kx - 3 = 0 has distinct real roots for all real values of k

We know that for a quadratic equation ax^2 + bx + c = 0, discriminant D = b^2 - 4ac
The condition for 'distinct' real roots is D > 0. Thus, to prove that x^2 + kx - 3 = 0 has distinct real roots, we got to prove D > 0 for it.

Lets have a look at what D comes out for x^2 + kx - 3 = 0

D
= k^2 - 4(1)(-3)
= k^2 + 12

We obviously understand that if k is a real number (and its given in the question that it is), k^2 will always be greater than or equal to 0 (k^2 ≥ 0). The reason of it being the fact that square of any real number can never be a negative number and thus it is always greater than or equal to 0.

Now, if k^2 is always greater than or equal to 0, then, by common sense, we can say that k^2 + 12 must be greater than or equal to 12. We can understand the same in form of a mathematical inequality. We know that 

If α, β be the roots of ax^2 + bx + c = 0 and γ, δ those of lx^2 + mx + n = 0, find the equation whose roots are (αγ + βδ) and (αδ + βγ)

Roots of the equation ax^2 + bx + c = 0 are α, β

Roots of the equation lx^2 + mx + n = 0 are γ, δ

By applying Sum and Product of roots formulae on both the equations, we will get,

α + β = -b/a          (1)

αβ = c/a                (2)

γ + δ = -m/l          (3)

γδ = n/l                 (4)

We have to find the equation whose roots are (αγ + βδ)  and (αδ + βγ). It has two roots, so obviously it is a quadratic equation. As we know, any quadratic equation can be written as x^2 - (sum of its roots) + product of its roots = 0. So, to find the equation which has roots (αγ + βδ)  and (αδ + βγ), we'll first have to find sum and product of those roots.

If alpha, beta be roots of equation x^2 - px + q = 0, then form an equation whose roots are q/(p - alpha) and q/(p - beta)

Okay. So lets begin with remarking this as a good question.

We are given the equation x^2 - px + q = 0 and it is said that its roots are α and β. We have to form an equation whose roots are q/(p - α) and q/(p - β). This equation which has to be found has two roots, so obviously it is also a quadratic equation.

The approach to these types of questions is very basic. First of all, we'll apply the Sum and Product of roots formulae on the given equation.

Sum of roots = -(-p)/1

=> α + β = p               (1)

Product of roots = q/1

=> αβ = q                   (2)

We know that any Quadratic Equation can be written in the form x^2 - (sum of its roots)x + product of its roots = 0. So, to find the equation whose roots are q/(p - α) and q/(p - β), we'll have to find the sum and product of these roots.

α, β are roots of x^2 + px + q = 0. Form an equation whose roots are (α + β)^2 and (α - β)^2.

Equation x^2 + px + q has roots α, β. By using the sum and product of roots formulae, we'll get

α + β = -p               (1)

αβ = q                    (2)

We have to find an equation who roots are (α + β)^2 and (α - β)^2.

We know that any Quadratic equation can be written as x^2 - (sum of its roots)x + product of its roots = 0. Hence to find the equation with roots (α + β)^2 and (α - β)^2, we'll have to find the sum and product of those roots.

Sum of roots

= (α + β)^2 + (α - β)^2

= (-p)^2 + α^2 + β^2 - 2αβ               (by Using (1))

= p^2 + (α + β)^2 - 2αβ - 2αβ        (by using a^2 + b^2 = (a + b)^2 - 2ab)

= p^2 + (α + β)^2 - 4αβ

= p^2 + (-p)^2 - 4αβ                         (by using (1))

= p^2 + p^2 - 4q                               (by using (2))

= 2p^2 - 4q

= 2(p^2 - 2q)

Equation x^2 + 3x + 1 = 0 has roots a, b. Find the equation whose roots are (i) a/b, b/a (ii) a^3, b^3 (iii) a/(2b+3), b/(2a+3)

We are given that the equation x^2 + 3x + 1 = 0 has roots a, b.

By using Sum and Product of roots formulae, we'll get

a + b = -3               (1)

a.b = 1                   (2)

Also, we know that any Quadratic equation can be written as

x^2 - (sum of its roots)x + (product of its roots) = 0

Keeping these in mind, we will try finding equations for the pairs of roots given in the question.

(i) We have to find the equation whose roots are a/b and b/a. We can easily get the equation if we calculate the sum and product of its roots.

Sum of roots

= a/b + b/a

= (a^2 + b^2)/ab

= [(a + b)^2 - 2ab] / ab

One root of equation ax^2 + bx + c = 0 is equal to nth power of the other root. Show that (a^n .c)^(1/n+1) + (a.c^n)^(1/n+1) + b = 0

The equation ax^2 + bx + c = 0 has roots such that one roots is equal to nth power of the other. We have to prove (a.c^n)^(1/ n+1) + (a^n .c)^(1/ n+1) + b = 0

We can assume the roots to be r and r^n. By applying sum and product of roots formulae on equation ax^2 + bx + c= 0, we'll get

Product of roots = c/a

=> r . r^n = c/a

=> r^(n+1)= c/a

=> r = (c/a)^(1/ n+1)

Sum of roots = -b/a

=> r + r^n = -b/a

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)

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.

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