[Math] Solving for x with exponents (algebra)

algebra-precalculusquadratics

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 $x$.

Thanks in advance for replies.

Best Answer

$8x+3=3x^2$

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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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Reference: http://www.teacherschoice.com.au/Maths_Library/Algebra/Alg_18.htm

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