1. ## Solving quadratic equation with surds

solve the equation x^2 - (6√3)x^2 + 24 = 0 giving the answer as surds, simplified as far as possible.

i can't isolate the '6√3' without getting something wrong and ruining the equation.

2. Hi there,

Try rearranging it into the form:

$(1-6\sqrt{3})x^2+24=0$

Can you take it from there?

3. Originally Posted by Miasmagasma
solve the equation x^2 - (6√3)x^2 + 24 = 0 giving the answer as surds, simplified as far as possible.

i can't isolate the '6√3' without getting something wrong and ruining the equation.
Are you sure this is the right equation? Is the second coeffiecent to the first degree?

4. I would suggest using the Quadratic Formula...

5. wow i'm even more confused now

you have to be gentle with me as i'm teaching myself this stuff.

i can't see how to get the equation to that point, stroodle. and i wouldn't know what to do from there anyway.

??

6. Originally Posted by 11rdc11
Are you sure this is the right equation? Is the second coeffiecent to the first degree?
i'm afraid i don't understand what you mean.

7. If you didn't mean to square the $x$ in the second term of the equation;

i.e. $x^2-(6\sqrt{3})x+24=0$

Then you could either use the quadratic formula or complete the square (as follows) to solve for $x$

$x^2-(6\sqrt{3})x+24=0$

$\left (x-\frac{6\sqrt{3}}{2}\right ) ^2-\frac{108}{4}+\frac{96}{4}=0$

$\left ( x-\frac{6\sqrt{3}}{2}+\frac{2\sqrt{3}}{2}\right )\left ( x-\frac{6\sqrt{3}}{2}-\frac{2\sqrt{3}}{2}\right )=0$

$(x-2\sqrt{3})(x-4\sqrt{3})=0$

So, $x=2\sqrt{3}$ or $x=4\sqrt{3}$

Hope that helps

8. Originally Posted by Miasmagasma
wow i'm even more confused now

you have to be gentle with me as i'm teaching myself this stuff.

i can't see how to get the equation to that point, stroodle. and i wouldn't know what to do from there anyway.

??
If the equation is $x^2-(6\sqrt{3})x^2+24=0$

Then you can rearrange it to the form:

$(1-6\sqrt{3})x^2+24=0$ ,as when you expand $(1-6\sqrt{3})x^2$ you get $x^2-(6\sqrt{3})x^2$

From there you can simply solve as follows:

$(1-6\sqrt{3})x^2+24=0$

$(1-6\sqrt{3})x^2=-24$

$x^2=\frac{-24}{1-6\sqrt{3}}$

$x=\pm\sqrt{\frac{24}{6\sqrt{3}-1}}$

9. Even if it's not, you can still use the quadratic formula anyway.

10. Originally Posted by Prove It
Even if it's not, you can still use the quadratic formula anyway.

as Prove it said you could use the quad equation letting

$A= (1-6\sqrt{3})$

$B= 0$

$C = 24$

11. Originally Posted by Stroodle
If you didn't mean to square the $x$ in the second term of the equation;

i.e. $x^2-(6\sqrt{3})x+24=0$

Then you could either use the quadratic formula or complete the square (as follows) to solve for $x$

$x^2-(6\sqrt{3})x+24=0$

$\left (x-\frac{6\sqrt{3}}{2}\right ) ^2-\frac{108}{4}+\frac{96}{4}=0$

$\left ( x-\frac{6\sqrt{3}}{2}+\frac{2\sqrt{3}}{2}\right )\left ( x-\frac{6\sqrt{3}}{2}-\frac{2\sqrt{3}}{2}\right )=0$

$(x-2\sqrt{3})(x-4\sqrt{3})=0$

So, $x=2\sqrt{3}$ or $x=4\sqrt{3}$

Hope that helps
i wish it did but i don't understand any of it. i have never used that quadratic formula. oh i think you're right... that x wasn't meant to be squared. really sorry to anyone who tried to do it with the wrong equation.

12. Originally Posted by Miasmagasma
i wish it did but i don't understand any of it. i have never used that quadratic formula. oh i think you're right... that x wasn't meant to be squared. really sorry to anyone who tried to do it with the wrong equation.
i now recognise the formula actually but i'm confused about the second step ( where the 2 brackets = 0 ) and the final step. can you walk me through like you were talking to a dimwit?

13. You should have a look here: Practical Algebra Lessons under the "intermediate algebra topics"; it explains quadratics a lot better than I'm able to.

For the method I used (completing the square) go to Completing the Square: Solving Quadratic Equations otherwise there's also a link there on how to use the quadratic formula - which is essentially the same thing as it's derived by completing the square - where you can just plug in values into the formula to solve for $x$

14. Yes you may use the quadratic formula as what Prove It said..

15. Originally Posted by Miasmagasma
solve the equation x^2 - (6√3)x^2 + 24 = 0 giving the answer as surds, simplified as far as possible.

i can't isolate the '6√3' without getting something wrong and ruining the equation.

Hi all, I am a new member of forum. Would a newcomer be warmly welcome here? Good day you guys!!!

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