Question

(x^1/2)² - (3√3)x^1/2 + 6 = 0... Please help, using quadratic formula

Answer 1

Bit nasty. Alright, well Im sure you know about how a quadratic formula can be used when you have something like ax^2 + bx + c. Well, even though this formula uses x to the first power and x to the second power, it can really be done with anything as long as you have the leading power of x be 2 times the middle power of x. For example, these are all candidates for quadratic formula:

\[x ^{6}+x ^{3}+1 \]
\[x^{5}-2 x^{\frac{ 5 }{ 2 }}-3\]
\[\sqrt{x}-\sqrt[4]{x}+1 \]

The reason all of these work is because the first power of x is double the second power of x and the 3rd term is a constant. So for yours, you have:
\[(x ^{\frac{ 1 }{ 2 }})^{2}-3\sqrt{3}x ^{\frac{ 1 }{ 2 }} +6=0 \]

Now that left (x^1/2)^2, we can multiply that 2 exponent in and make it just be x^1. So this means you do have it in quadratic form because the first power of x is 2 times the 2nd power of x. So now we'll just plug numbers into the quadratic formula. If you recall, quadratic formula is:

\[\frac{ -b \pm \sqrt{b ^{2}-4ac}}{ 2a } \]So if I plug those numbers in:

\[\frac{ 3\sqrt{3}\pm \sqrt{(3\sqrt{3})^{2}-4(1)(6)} }{ 2 } \]

Now the (3sqrt3)^2 is the same as 3^2 times sqrt(3)^2. The power of 2 distributes basically. So this means (3sqrt3)^2 = 9 * 3 = 27. The -(4)(1)(6) becomes 24. So inside of the radical, we have 27-24 = 3. So now we can write things like this:

\[\frac{ 3\sqrt{3}\pm \sqrt{3} }{ 2 }\implies \frac{ 3\sqrt{3} }{ 2 }\pm \frac{ \sqrt{3} }{ 2 } \]

So now I can solve this for the plus and for the minus. The plus version of the plus or minus gives us:

\[\frac{ 3\sqrt{3} }{ 2 }+\frac{ \sqrt{3} }{ 2 }=\frac{ 4\sqrt{3} }{ 2 }=2\sqrt{3}\]And for the minus version we get:

\[\frac{ 3\sqrt{3} }{ 2 }-\frac{ \sqrt{3} }{ 2 }=\frac{ 2\sqrt{3} }{ 2 }=\sqrt{3} \]

So these are our two answers. Careful now, though. Normally we would say x = 2sqrt(3) and sqrt(3), but that is not true this time. Recall our original equation, it was not in a form like ax^2 + bx + c, but actually in the form of ax^1 + bx^1/2 + c. So this means instead of saying x = sqrt3 and 2sqrt3, we have to say:
\[\sqrt{x}=2\sqrt{3}, \sqrt{3} \]

I know this is quite bit, so I hope it makes sense.

Answer 2

@javydiaz97

Answer 3

Thank you, I understand know. however I have one question.

How did 3√3 + or - √3 / 2 ------ go to ------- 4√3/2 and then 2√3

Answer 4

Its reduced.

\[\frac{ 4\sqrt{3} }{ 2}=\frac{ 4 }{ 2 }*\frac{ \sqrt{3} }{ 1 }\]
\[\frac{ 4 }{ 2 }*\frac{ \sqrt{3} }{ 1 }=\frac{ 2 }{ 1 }*\frac{ \sqrt{3} }{ 1 } \] 4 over 2 just reduced to 2 over 1.

Answer 5

Okay sorry but where is the 4 coming from exactly?

Answer 6

Well, since we have the +/- sign, we have to do two equations. We have to do one where the sign is plus and one where the sign is minus. It's like a combination of like terms. If I have 3x + x, it gives me 4x. Well, we can think of that sqrt(3) as an x.

\[\frac{ 3\sqrt{3} }{ 2 }+\frac{ \sqrt{3} }{ 2 } -> \frac{ 3x }{ 2 }+\frac{ x }{ 2 } \]So its basically like doing 3x + x and then 3x - x, with the 2 in the denominator just coming along.

Answer 7

Wow, thank you very much.
I appreciate the help.

Answer 8

Yeah, np ^_^