The roots of the equation 2x^2-3x+5=0 are alpha and beta and the roots of the equation px^2+x+q=0 are alpha-1 and beta-1. Find the value of p and q

2x^2 - 3x + 5 = 0 has roots alpha, beta

alpha + beta = -(-3)/2
=> alpha + beta = 3/2

(alpha)(beta) = 5/2

px^2 + x + q = 0 has roots (alpha - 1) and (beta - 1). Hence,

(alpha - 1) + (beta - 1) = -1/p
=> alpha + beta - 2 = -1/p

Since we have found alpha + beta = 3/2 above,

=> 3/2 - 2 = -1/p
=> -1/2 = -1/p
=> p = 2

Also, 

(alpha - 1)(beta - 1) = q/p
=> (alpha)(beta) - (alpha) - (beta) + 1 = q/p
=> (alpha)(beta) - (alpha + beta) + 1 = q/p

Using (alpha)(beta) = 5/2, alpha + beta = 3/2, and p = 2

=> 5/2 - 3/2 + 1 = q/2
=> 2 = q/2
=> q = 4

Hence, p = 2 and q = 4.

Find the condition that one root of the equation ax^2+bx+c=0 is greater than the other by h.

We have to find the condition for which roots of ax^2 + bx + c = 0 have a difference of h.

We know that
Difference of roots = √(b²-4ac) / a or (√D)/a

=> h = √(b²-4ac)

=> h² = b²-4ac

When h² = b²-4ac holds true, the equation ax^2 + bx + c = 0 will have one root greater than the other by h.

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 α, β are the roots of 3x^2 +5x-1=0, Construct equations whose root are (i) 5α, 5β (ii) α^2, β^2 (iii) 1/α, 1/β (iv) α+1/β, β+1/α

It's given that α, β are the roots of equation 3x^2 + 5x - 1 = 0. By applying the sum and product of roots formulae, we'll get

α + β = -5/3
αβ = -1/3

We also know that any quadratic equation is actually x^2 - (sum of its roots)x + (product of its roots) = 0. Like here, we have the equation x^2 -(-5/3)x + (-1/3) = 0 => 3x^2 + 5x - 1 = 0

(i) We have to find the equation whose roots are 5α, 5β

Sum of these roots = 5α + 5β = 5(α + β) = 5(-5/3) = -25/3
Product of these roots = 5α . 5β = 25αβ = -25/3

Hence, the equation will be

x^2 - (sum of roots)x + (product of roots) = 0
=> x^2 -(-25/3)x + (-25/3) = 0
=> 3x^2 + 25x - 25 = 0

(ii) The roots are α^2, β^2

Their sum = α^2 + β^2
= (α + β)^2 - 2αβ
= (-5/3)^2 - 2(-1/3)
= 25/9 + 2/3
= 31/9

Their Product = α^2.β^2
= (αβ)^2
= 1/9

Hence, the equation:
x^2 - (31/9)x + 1/9 = 0
=> 9x^2 -31x + 1 = 0

(iii) The roots are 1/α, 1/β

Their sum = 1/α +  1/β
= (α + β)/αβ
= (-5/3)/(-1/3) = 5

Their product = (1/α)(1/β) = 1/αβ = -3

Hence, the equation:
x^2 - (5)x + (-3) = 0
=> x^2 -5x - 3 = 0

(iv) The roots are (α + 1/β), (β + 1/α)

Their sum
= (α + 1/β)+ (β + 1/α)

= α + β + 1/α + 1/β

= (α + β) + (α + β)/αβ

= -5/3  + (-5/3)/(-1/3)

= -5/3 + 5

= 10/3

Their Product

= (α + 1/β) (β + 1/α)

= αβ + 1 + 1 + 1/αβ

= -1/3 + 2  +(-3)

= -4/3

Hence, the equation:

x^2 -(-10/3)x + (-4/3) = 0

=> 3x^2 + 10x - 4 = 0

Two students are solving a quadratic equation of form x^2 + px + q = 0. One starts with a wrong value of p and finds the roots to be 2 and 6. The other starts with a wrong value of q and finds the roots to be 2 and -9. Find the correct roots of the equation.

We are told that the equation is of the form x^2 + px + q = 0.

The first guy, who has got the value of p wrong, solves the equation and finds out the roots to be 2 and 6. 

To understand the case in an easier and better way, lets assume that he thought the equation was x^2 + p'x + q = 0. The roots of this equation are 2 and 6. We can now deduce that in this case

Sum of roots = -p'/1 = 2 + 6 

=> p' = -8

Product of roots = q/1 = 12

=> q = 12

This gives us the value of q, which is 12. The first guy had got the value of the coefficient of p wrong, but he got the value of q right, which means that q = 12.

Similarly. the second guy, who has the wrong value of q, solves the equation, say, x^2 + px + q' = 0, and finds out the roots to be 2 and -9. So, we can say that

Sum of roots = -p = 2 + (-9) = -7 => p = 7

Product of roots = q' = 2 * -9 = -18

This guy has got the value of p right, so p = 7.

Hence, the original equation must be x^2 + 7x + 12 = 0, solving which we'd get the roots -4 and -3.

ax^2 + bx + c = 0 and bx^2 + cx + a = 0 have a common root. Prove that a+b+c=0 or a=b=c

We are given the equations ax^2 + bx + c = 0 and bx^2 + cx + a = 0 and we are told that they have a common root. Lets assume the common root to be t.

Since it is a common root, it should satisfy both these equations. So, we'll have

at^2 + bt + c = 0
bt^2 + ct + a = 0

By substracting the second equation from the first,

at^2 + bt + c - bt^2 - ct - a = 0

Now, we'll pair the terms as indicated by the colours

=> at^2 - a - bt^2 + bt - ct + c = 0
=> a(t^2 - 1) - bt(t - 1) - c(t - 1) = 0

We can take (t - 1) common from the expression, because t^2 - 1 = (t + 1)(t - 1)

=> (t - 1) [ a(t + 1) - bt - c] = 0
=> t - 1 = 0, or a(t + 1) - bt - c =0

From t - 1 = 0, we get t = 1. This gives us that the common root is 1 and substituting it in either of the equation will give a + b + c = 0

We'll go on with a(t + 1) - bt - c = 0
 =>at - bt = c - a
=> t = (c - a)/(a - b)

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 ax^2 + bx + c = 0 has equal roots, then c = ?

We have the equation ax^2 + bx + c = 0 and it is given that the equation has equal roots.

If the equation has equal roots, it can be deduced that the difference of roots for the equation would be 0. For eg, if an equation has roots 3 and 3 (equal roots), the difference of roots will be 0. So, we'll apply this condition on the difference of roots formula

Difference of roots = √D / a  = √(b²-4ac) / a
=> 0 = √(b²-4ac) / a
=> 0 = √(b²-4ac)
=> b² - 4ac = 0
=> 4ac = b^2
=> c = b^2/4a

Hence, if ax^2 + bx + c = 0 has equal roots, then c = b^2/4a

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.

If roots of equation ax^2 + bx + c = 0 are alpha and beta, Find Alpha^3 + Beta^3.

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

Sum of roots = -b/a

Product of roots = c/a

Here, the roots are alpha and beta, so 

α + β  = -b/a

αβ = c/a

Now, we'll find α^3 + β^3 by using the formula a^3 + b^3 = (a + b)(a^2 - ab + b^2). We'll keep substituting the α + β and αβ with the results we found.

α^3 + β^3

=> (α + β)(α^2 - αβ + β^2)

=> -b/a (α^2 - αβ + β^2 + 3αβ - 3αβ)

(This addition and subtraction of 3αβ makes sure that the expression is not altered and we get to simplify terms to make (a + b)^2 form)   

=> -b/a (α^2 + 2αβ + β^2 - 3αβ)

=> -b/a ((α + β)^2 - 3αβ)

=> -b/a ((-b/a)^2 - 3(c/a))

=> -b/a (b^2/a^2 - 3c/a)

=> -b/a (b^2 - 3ac)/a^2

=> (3abc - b^3)/a^3

 (3abc - b^3)/a^3 will be the answer.

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

(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.

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