# Solving for x with exponents (algebra)

So I am trying to help a friend do her homework and I am a bit stuck.

$$8x+3 = 3x^2$$

I can look at this and see that the answer is $3$, but I am having a hard time remembering how to solve for $x$ in this situation.

Could someone be so kind as to break down the steps in solving for $x$.

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Thank you all for your comments it would seem i didnt give enough information from the start and for that i apologize for wasting your time. I will type the question as it is on her work so you can see what i am trying to get at. the problem above was simplified from the original. "GIVEN: M is between l and n; lm=2x+1; mn=6x-3; ln = 3x^2-5 find the length of each segment. this is what i am trying to show the work for. I can see the answer just have no idea how to express the work. –  Michael Cole Sep 28 '12 at 7:47

$8x+3=3x^2$

You can solve the above equation using "Splitting middle term of quadratic equation" formula as well. It is mostly useful for simple equations like above.

Solution:

You can re-write the above as:

$3 \cdot x^2 - 8 \cdot x -3 = 0$

Now try to express the middle term: $8 \cdot x$ as the factor of the product of the coefficient of the other two terms: $3 \cdot (-3) = (-9)$

Now $9$ can be represent as $9 \cdot 1$

Now look at the sign of the middle term: $8 \cdot x$ and it is $-$ (negetive)

Since it is negetive, express $-8x$ as $(-9x + 1x)$

So, the equation in this case will be:

$3 \cdot x^2 -9 \cdot x + 1 \cdot x -3 = 0$

$\implies (x-3)(3x+1) = 0$

$\implies x =3, -\frac{1}{3}$

Now x can't be negative.

So, $x = 3$

Now put x =3 in the above mentioned equations:

So,

$lm = 2 \cdot x +1 = 7$

$mn = 6 \cdot x -3 = 15$ and

$ln = 3 \cdot x^2 -5 = 22$


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This is a quadratic equation: the highest power of the unknown is $2$. Rearrange it to bring everything to one side of the equation:

$$3x^2-8x-3=0\;.$$

If you can easily factor the resulting expression, you can take a shortcut, but otherwise you either complete the square or use the quadratic formula.

Completing the square relies on the fact that $(x+a)^2=x^2+2ax+a^2$. First factor out the coefficient of $x^2$:

$$3x^2-8x-3=3\left(x^2-\frac83x-1\right)\;.\tag{1}$$

Now notice that if you set $a=-\dfrac{8/3}2=-\dfrac43$, you’ll have $$(x+a)^2=\left(x-\frac43\right)^2=x^2-\frac83x+\frac{16}9\;,$$ which agrees in all but the constant term with the expression in parentheses in $(1)$. Thus,

$$x^2-\frac83x-1=\left(x-\frac43\right)^2-\frac{16}9-1=\left(x-\frac43\right)^2-\frac{25}9\;,$$

and on substituting back into $(1)$ we have

$$0=3x^2-8x-3=3\left(x-\frac43\right)^2-\frac{25}3\;.$$ Rearranging this gives us

$$3\left(x-\frac43\right)^2=\frac{25}3\;,$$ or $$\left(x-\frac43\right)^2=\frac{25}9\;.$$ Finally, taking the square root on both sides and remembering that there are two square roots, one positive and one negative, we get

$$x-\frac43=\pm\frac53$$ and therefore $$x=\frac53+\frac43=\frac93=3\quad\text{or}\quad x=-\frac53+\frac43=-\frac13\;.$$

The quadratic formula can be derived by applying the method of completing the square to the general quadratic equation $ax^2+bx+c=0$. The result is that

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

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Here's a way without the formula. Using "completing the square"

\begin{align*} 3x^2-8x-3&=0\\ x^2-\frac{8}{3}x-1&=0\\ \left(x-\frac{4}{3}\right)^2-\frac{16}{9}-1&=0\\ x-\frac{4}{3}&=\pm \sqrt{\frac{25}{9}}\\ x &= \frac{4}{3}\pm \frac{5}{3}\\ x = \frac{9}{3} = 3\ \ &\mathrm{or}\ \ x = -\frac{1}{3}. \end{align*}

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Use the quadratic formula. First, move all the terms to one side: $8x+3=3x^2 \Rightarrow 3x^2-8x-3=0$. Then apply the formula

$\Large \hspace{60mm}x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$,

where $a=3, b=-8,$ and $c=-3$.

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Rearranging the equation (bringing all the terms onto one side) we have:

$$3x^2 - 8x - 3 = 0.$$

Now there are (at least) two options.

1. Factorise the quadratic on the left. To make it easier, you can divide both sides of the equation by $3$ to get

$$x^2 - \frac{8}{3}x - 1 = 0$$

in which case you need to factorise $x^2 - \frac{8}{3}x - 1$, that is, write it as $(x - a)(x - b)$ for some $a$ and $b$. The way we determine $a$ and $b$ is by looking at the coefficent of $x$, which is always the negative of the sum of $a$ and $b$ (i.e. $-\frac{8}{3} = -(a + b)$), and by looking at the constant term, which is always the product of $a$ and $b$ (i.e. $-1 = ab$). Solving these two equation for $a$ and $b$, we find the factorisation $x^2 - \frac{8}{3}x - 1 = (x - 3)(x + \frac{1}{3})$, so our equation now reads

$$(x - 3)\left(x + \frac{1}{3}\right) = 0.$$

Now the null factor law states that if the product of two things is zero, then at least one of them is zero. In this case, that means that either $x - 3 = 0$ (i.e. $x = 3$), or $x + \frac{1}{3} = 0$ (i.e. $x = -\frac{1}{3}$).

2. Use the quadratic formula which gives:

$$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{8\pm\sqrt{(-8)^2-4(3)(-3)}}{2(3)} = \frac{8\pm\sqrt{100}}{6} = \frac{8\pm 10}{6},$$

so $x = \frac{8 + 10}{6} = 3$ or $x = \frac{8 - 10}{6} = -\frac{1}{3}$.

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Rewrite the equation as $3x^2-8x-3=0$ and then use the quadratic formula with $a=3,b=-8,c=-3$. $$x=\frac{8\pm\sqrt{64+4*3*3}}{2*3} = \frac{8\pm 10}{6} = 3,-\frac{1}{3}$$

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2*3 = 6... not 8 –  Alex Sep 28 '12 at 7:24
Sorry everyone. Made an error an forgot to check my work. It's fixed now. –  chris Sep 28 '12 at 16:34