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.

If roots of equation (h^2-a^2)x^2-2hkx+(k^2-b^2)=0 are real and equal, prove that h^2/a^2 + k^2/b^2 = 1

We have the equation (h^2-a^2)x^2-2hkx+(k^2-b^2)=0
It is given that the roots of this equation are real and equal, hence D should be equal to 0

D= 0
=> b^2 - 4ac = 0
=> (-2hk)^2 - 4(h^2 - a^2)(k^2 - b^2) = 0
=> 4h^2.k^2 -4(h^2.k^2 - h^2.b^2 - a^2.k^2 + a^2.b^2) = 0
=> 4h^2.k^2 -4h^2.k^2 + 4h^2.b^2 + 4a^2.k^2 - 4a^2.b^2 = 0
=> 4h^2.b^2 + 4a^2.k^2 - 4a^2.b^2 = 0
=> h^2.b^2 + a^2.k^2 - a^2.b^2 = 0

Now, by dividing the equation by a^2.b^2, we'll get

=> h^2.b^2/a^2.b^2 + a^2.k^2/a^2.b^2 - a^2.b^2/a^2.b^2 = 0/a^2.b^2
=> h^2/a^2 + k^2/b^2 - 1 = 0
=> h^2/a^2 + k^2/b^2 = 1 

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.

If a^2 + b^2 + c^2 = 2a - 2b - 2, Find the value of 3a - 2b + c.

We are given that 

a^2 + b^2 + c^2 = 2a - 2b - 2
=> a^2 + b^2 + c^2 - 2a + 2b + 2 = 0
=> a^2 - 2a +1 + b^2 + 2b + 1 + c^2  = 0
=> (a - 1)^2 + (b + 1)^2 + c^2 = 0

(a - 1)^2, (b + 1)^2, and c^2 are all squares, which means all are >/= 0. As the minimum value of each of them individually is 0, their sum could only be 0 when all of them become equal to 0.

(a - 1)^2 is equal to 0 at a = 1
(b + 1)^2 is equal to 0 at b = -1
c^2 is equal to 0 at c = 0

Hence, a = 1, b = -1, and c = 0.

Which means 3a - 2b + c = 3(1) - 2(-1) + 0 = 5