algebra-precalculus
I've been working at this problem for a while and I'm having troubled, with the exponents.
I would appreciate if someone could teach me how to solve this for an exam tomorrow, this is one of few problems that I struggle with thanks in advance, I am solving for x.
$8x+3=3x^2$
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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Reference: http://www.teacherschoice.com.au/Maths_Library/Algebra/Alg_18.htm
By inspection, you could notice that, if $$f(x)=(1-x)^{5/2}-(5/3)(1-x)^{3/2}+1/3$$ $f(-1)=\frac{1}{3}+\frac{2 \sqrt{2}}{3}$ which is positive, $f(0)=-\frac{1}{3}$ which is negative and $f(1)=\frac{1}{3}$ which is positive. Then, the equation shows at least two roots, one between $-1$ and $0$, and another one between $0$ and $+1$. You could even go further and notice that the roots are close to $-\frac{1}{2}$ and $\frac{1}{2}$ since $f(-\frac{1}{2})=\frac{1}{24} \left(8-3 \sqrt{6}\right)=0.0271471$ and $f(\frac{1}{2})=\frac{1}{24} \left(8-7 \sqrt{2}\right)=-0.0791456$.
As mentioned in other answers and comments, getting accurate solutions will require purely numerical methods (these are quite simple is you know derivatives).
Let me know if you want me to elaborate in this direction.
Best Answer
Using hit and trial for small numbers you can see that $x=2$ is a root , hence $$ 4x^2(x-2)-9(x-2)=0$$
and then
$$(4x^2-9)(x-2)=0$$
$$(2x-3)(2x+3)(x-2)=0$$
Roots are $x = -1.5 , 1.5$ and $2 $