1. ## Self-learner having trouble with quadratic equation

Hi. First of all - sincere apologies if this is in the wrong place. I have read the rules, but unfortunately don't really know all categories of mathematics, or what comes under which umbrella.

Anyway, simple quadratic equations I don't find difficult.

However, this problem is seriously getting to me. The exercise book I have bought claims the answer to be 0.2, but I simply cannot find out why. I thought QE were meant to have ± values, and this is just making me crazy.

(x-1)^2 = x(x-3)

I have expanded the brackets to find:

X^2-2x+1 = X^2-3x

But I have honestly no clue where to go from next. I have tried researching it, or finding other similar problems online but to no avail.

2. This is a linear equation rather than quadratic. You have x^2 on both sides and it will cancel.

If you take x^2-3x from both sides you will get $x^2-2x+1 - x^2+3x = 0$ which simplifies to $x+1=0$

It follows that $x=-1$ is the solution because $(-1-1)^2 = -1(-1-3) \rightarrow 4 = 4$ whereas $x=0.2$ gives $(0.2-1)^2 = -0.2(0.2-3) \rightarrow 0.64 \neq 0.56$

edit: as a background you can have one solution to a quadratic equation (technically 2 equal roots). This happens when the discriminant $\left(\Delta = b^2-4ac\right)$ is equal to 0.

For example find the solution(s) to $4x^2-4x+1=0$ (answer in spoiler)

Spoiler:
$4x^2-4x+1 = (2x+1)^2 = 0 \rightarrow \: x = -\dfrac{1}{2}$

3. Thanks very much. It's in the QE part of my workbook... and the answer is listed as 0.2. I had solved it as you did before, but just assumed that was too simple or something.

I'll assume it's a misprint - thanks for your help.

4. Originally Posted by S1919
(x-1)^2 = x(x-3)
Looks like that equation contains a typo: should be (x - 1)^2 = x(x + 3) ?
That would explain the .2 solution:
x^2 - 2x + 1 = x^2 + 3x
5x = 1
x = 1/5 = .2