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.

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 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

(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