# Math Help - 2 questions on my math homework

1. ## 2 questions on my math homework

I can't seem to figure these two questions out. A little assistance please?

1)

x^2/3 + 2x^1/3 - 63 = 0

The solution set is ( , )

2)

For each given pair of numbers find a quadratic equation with integral coefficients that has the numbers as its solutions.

-4i, 4i

2. Originally Posted by knowledgekick
I can't seem to figure these two questions out. A little assistance please?

1)

x^2/3 + 2x^1/3 - 63 = 0

The solution set is ( , )

2)

For each given pair of numbers find a quadratic equation with integral coefficients that has the numbers as its solutions.

-4i, 4i
Is your first question $x^{\frac{2}{3}} + 2x^{\frac{1}{3}} - 63 = 0$???

3. Originally Posted by harish21
Is your first question $x^{\frac{2}{3}} + 2x^{\frac{1}{3}} - 63 = 0$???
Yes, thank you.

4. Originally Posted by knowledgekick
Yes, thank you.
Substitute $u = x^{1/3}$ on the LHS:

${u^2} + 2u - 63 = 0$

$u^2+ 2u = 63$

$u^2+2u+1 = 64$

$(u+1)^2 = 64$

Take the square root of both sides:

$u+1 = \pm 8$

$u = -9$ OR $u = 7$

Substitute back for $u = x^{1/3}$

$x^{\frac{1}{3}}= -9$ OR $x^{\frac{1}{3}} = 7$

Cube both sides:

$x = -729$ OR $x = 343$

Now try plugging these values into your question and check!

5. Yes! Thank you very much! I'm clueless on what to do for the other problem though. Any idea?

6. Originally Posted by knowledgekick
Yes! Thank you very much! I'm clueless on what to do for the other problem though. Any idea?
Your solutions are given as $4i$ and $-4i$

So, you can set up the equation as:

$(x-4i)(x+4i)=0$

Remember, for imaginary number $i$,

$i^{2} = -1$

7. I got x^2 + 16 for that. I think it's right. Thanks a lot!

I have a couple more that are bothering me now. If you could help that would be great. If not, I understand...

1)

x^2/3 + 3x^1/3 - 28 = 0

Solution set is ( , )

I tried doing this the same way you did the first one but I'm not getting it!

2)

Find the vertex

a= x^2 + 18x - 8

8. Originally Posted by knowledgekick
I got x^2 + 16 for that. I think it's right. Thanks a lot!

I have a couple more that are bothering me now. If you could help that would be great. If not, I understand...

1)

x^2/3 + 3x^1/3 - 28 = 0

Solution set is ( , )

I tried doing this the same way you did the first one but I'm not getting it!

2)

Find the vertex

a= x^2 + 18x - 8
$x^{2/3} + 3x^{1/3} - 28 = 0$

Substitute $u = x^{1/3}$ on the LHS:

You have

$u^{2} + 3u -28 = 0$

$u^{2} +7u -4u -28 = 0$

$u(u+7)-4(u+7) = 0$

$(u+7)(u-4) = 0$

therefore $u = -7$ OR $u = 4$

substitute $u = x^{1/3}$ and solve as done before

9. Originally Posted by knowledgekick
I got x^2 + 16 for that. I think it's right. Thanks a lot!

I have a couple more that are bothering me now. If you could help that would be great. If not, I understand...

1)

x^2/3 + 3x^1/3 - 28 = 0

Solution set is ( , )

I tried doing this the same way you did the first one but I'm not getting it!

2)

Find the vertex

a= x^2 + 18x - 8
(2)

$a= x^2 + 18x - 8$ is a parabola..........(1)

The equation of a parabola can be written as:

$y = ax^{2}+bx+c$.........(2)

And the vertex of the parabola is given by :vertex = $\frac{-b}{2a}$

Compare equation (1) and equation (2), your values are:

$b = 18$ and $a = 1$