1. ## Determinants Problem

Hey guys,

I got into quite a predicament lately when I tried to solve this problem. We need to find the value for x such that the determinants below are equal. I tried to expand then equate but got a strange equation, that I couldn't solve. I tried row operations but that failed too. Can you guys please help? Thanks
Yes, the one on the left is 2*2 and on the right is 3*3.
Thanks again.

2. I don't understand how the equation you got was strange, you should have included some working so someone could spot where it might have goine wrong.

Expanding both sides you should end up with

$\displaystyle x(1-x) + 3 = x(x-5) -18 +3x + 18$

Which is just a quadratic in x. Bring everything onto one side and use the quadratic formula.

Hope this helps.

pomp.

Hey pomp,

Thanks for the quick response, my question lyes in the quadratic equation itself. When I solved it I got
2x^2 - 3x - 3 = 0
and when I tried to solve it by factorization, I couldn't end up with any solution.
The same thing happened with the quadratic equation (ending up with a negative sign under the root). I'll really appreciate the help. Thanks again.

4. Originally Posted by Zakaria007
Hey pomp,

Thanks for the quick response, my question lyes in the quadratic equation itself. When I solved it I got
2x^2 - 3x - 3 = 0
and when I tried to solve it by factorization, I couldn't end up with any solution.
The same thing happened with the quadratic equation (ending up with a negative sign under the root). I'll really appreciate the help. Thanks again.
$\displaystyle f(x)=2x^2 - 3x - 3$ has two real roots so you made a mistake by using the formula $\displaystyle x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$. Precisely $\displaystyle b^2-4ac=33$ here, which is not a negative number.
Good luck.

5. ## Final result

OK man, I get my mistake. I RUSHED!
The real solution is 9. Thanks guys.

6. Sorry but that is incorrect. Arbolis pointed out that the equation has 2 real roots.

$\displaystyle f(x) = 2x^2 - 3x -3 = 0$

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here we have $\displaystyle a = 2$ $\displaystyle b = -3$ $\displaystyle c = -3$ so

$\displaystyle x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \times 2 \times (-3)}}{2 \times 2}$

$\displaystyle x = \frac{3 \pm \sqrt{9 + 24}}{4}$

$\displaystyle x = \frac{3 + \sqrt{33}}{4}$

and

$\displaystyle x = \frac{3 - \sqrt{33}}{4}$