1. ## Questions!

Solve for x:

a) 4x^2 + 3x - 2 = 0
b) 4x + cuberoot x - 2 = 0. (Hint: rename cuberoot x = z)
c)[(x + 4) / (x - 2)] - [(x - 3) / (x - 4)] = 0.
d) 3x = squareroot(3 + x)

2. Originally Posted by princess_anna57
a) 4x^2 + 3x - 2 = 0
This is a quadratic. If you don't know what else to do with it, use the quadratic formula. As this doesn't factor, that's what I'm going to do.
$4x^2 + 3x - 2 = 0$

So....
$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 4 \cdot (-2)}}{2 \cdot 4}$

$x = \frac{-3 \pm \sqrt{9 + 32}}{8}$

$x = \frac{-3 \pm \sqrt{41}}{8}$

-Dan

3. Originally Posted by princess_anna57
b) 4x + cuberoot x - 2 = 0. (Hint: rename cuberoot x = z)
$4x + \sqrt[3]{x} - 2 = 0$

Why were you given this problem? It seems above the level that I thought you were at.

Let $z = \sqrt[3]{x}$, then $x = z^3$.

Thus
$4z^3 + z - 2 = 0$

This has no rational factors, so the only way to do this (neglecting Cardano's method, which I'll do if you really want me to) is to use a graphing calculator to estimate it, or numerically calculate it.

I get that
$z \approx 0.689398, -0.344699 \pm 0.778751 \cdot i$

-Dan

4. Originally Posted by princess_anna57
c)[(x + 4) / (x - 2)] - [(x - 3) / (x - 4)] = 0.
$\frac{x + 4}{x - 2} - \frac{x - 3}{x - 4} = 0$

First note that the solution set cannot contain 2 or 4, since that is outside the domain of the original problem.

Now add the fractions by getting a common denominator:
$\frac{x + 4}{x - 2} \cdot \frac{x - 4}{x - 4} - \frac{x - 3}{x - 4} \cdot \frac{x - 2}{x - 2} = 0$

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

$\frac{(x^2 - 16) - (x^2 - 5x + 6)}{(x - 2)(x - 4)} = 0$

$\frac{5x - 22}{(x - 2)(x - 4)} = 0$

The only way the LHS can be 0 is for the numerator be 0. Thus set
$5x - 22 = 0$

$x = \frac{22}{5}$

(Note that x is neither 2 or 4, so this is a valid solution.)

-Dan

5. Originally Posted by princess_anna57
d) 3x = squareroot(3 + x)
$3x = \sqrt{3 + x}$

Square both sides:
$9x^2 = 3 + x$

$9x^2 - x - 3 = 0$

Again this is a quadratic that doesn't factor. So use the quadratic formula. I got:
$x = \frac{1 \pm \sqrt{109}}{18}$

Now, we need to check each of these in the original equation because squaring both sides tends to add spurious solutions. In this case we may simply note that if we use the "-" sign for x we don't have a solution. (I used the decimal approximation of the two x values to check in the original equation.)

So the solution is:
$x = \frac{1 + \sqrt{109}}{18}$

-Dan

6. With b), can you show me how to do the equation if it wasn't a cuberoot ie 4x + 3 squarerootx - 2 = 0?
Thank you so much for answering, I really appreciate it.

7. Originally Posted by princess_anna57
With b), can you show me how to do the equation if it wasn't a cuberoot ie 4x + 3 squarerootx - 2 = 0?
Thank you so much for answering, I really appreciate it.
$4x + 3 \sqrt{x} - 2 = 0$

Let $z = \sqrt{x}$, that is to say let $x = z^2$.

Then
$4z^2 + 3z - 2 = 0$

This is, in fact, the equation from part a), which I already solved. So we know that
$z = \frac{-3 \pm \sqrt{41}}{8}$

So
$x = z^2 = \left ( \frac{-3 \pm \sqrt{41}}{8} \right ) ^2$

$x = \frac{9 \mp 6\sqrt{41} + 41}{64} = \frac{50 \pm 6\sqrt{41}}{64}$ <-- We don't really care which order the + or - comes in.

$x = \frac{25 \pm 3 \sqrt{41}}{32}$

-Dan

8. For d), how did you get 109 when you used the quadratic formula?
Many thanks.

9. Originally Posted by princess_anna57
For d), how did you get 109 when you used the quadratic formula?
Many thanks.

For a quadratic equation of the form $ax^2 + bx + c = 0$

the roots of the equation are given by $x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$

For $9x^2 - x - 3 = 0$, $a = 9$, $b = -1$ and $c = -3$

so under the square root we get: $b^2 - 4ac = (-1)^2 - (4)(9)(-3) = 109$

10. ## Thanks Jhevon

Thanks Jhevon, I knew the formula, I just put the wrong numbers in the wrong place. Thanks again.