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# The Sum And Product of Roots of Quadratics watch

1. Someone help me with this...

The equation kx^2 - (1 + k)x + (3k + 2) = 0 is such that the sum of its roots is twice their product. Find k and the two roots.
2. All quadratic equations can be written in the form x^2 + bx + c = 0 right? And let's say that quadratic has 2 solutions, p and q, then

(x - p)(x - q) = 0 = x^2 + bx + c

Expand the left hand side and compare it with the right hand side. What does b equal? What does c equal? So what can you say about the sum and product of the roots of a quadratic?
3. (Original post by Swayum)
All quadratic equations can be written in the form x^2 + bx + c = 0 right? And let's say that quadratic has 2 solutions, p and q, then

(x - p)(x - q) = 0 = x^2 + bx + c

Expand the left hand side and compare it with the right hand side. What does b equal? What does c equal? So what can you say about the sum and product of the roots of a quadratic?
I see that b = p + q and a = pq, but cannot understand how to relate this to the initial equation to find the roots.
4. b = -(p+q) and c = pq I think you mean.

So you know that, for quadratics, the coefficient of x is the negative sum of the roots and the constant term is the product (if the coefficient of x^2 is 1).

And you also know that "sum of its roots is twice their product".

What equation can you write down?
5. Ok, thanks a lot, found a now

a = -(p+q) = 2 (a-2) = 2pq

a = 2a - 4

4 = a
6. (Original post by Swayum)
b = -(p+q) and c = pq I think you mean.

So you know that, for quadratics, the coefficient of x is the negative sum of the roots and the constant term is the product (if the coefficient of x^2 is 1).

And you also know that "sum of its roots is twice their product".

What equation can you write down?
actually i still do not get it...in my previous answer i assumed that the coefficient of x^2 (being a) will be equal to -(p + q).

However, if b = -6 = -(p+q) and is twice that of the product: Pq = a -2

then -12 = a -2
and - 10 = a

7. This looks like an MEI FP1 question. MEI expects you to know 'sum of roots = -b/a' and 'product of roots = c/a' for ax^2 + bx + c = 0. So you construct an equation in k which is easy to solve.

Plug that value for k into the original equation and you get a quadratic in x which factorises nicely. Clue - coefficients happen to sum to zero by chance. Solve for x and check that indeed sum = twice product.

Swayum is quite right in what he says. The non-MEI way would be to divide through by the coefficient of x^2 first and work with a quadratic with just one x^2.

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