I am trying to help my daughter on her math homework and I am having some trouble on some equation solving steps. My current major concern relies on understanding why $\sqrt{8}/2$ equal to $\sqrt{2}$.
Thanks in advance
Radicals – Why is ?8/2 Equal to ?2?
algebra-precalculusarithmeticradicals
Related Solutions
You and every other answer thus far have talked about starting by squaring both sides, which works, but there is a slightly easier path to take. Rewrite your equation so that there is only one square root on each side, for example: $$4\sqrt{x-3}=3+\sqrt{6x-17}$$ Now, when you square both sides, you'll still have another square root to deal with, but it's not the square root of a product, like the $\sqrt{(x-3)(6x-17)}$ you would have had: $$16(x-3)=9+6\sqrt{6x-17}+(6x-17)$$ Expand, collect like terms, and rearrange to get the square root by itself: $$5x-20=3\sqrt{6x-17}$$ Square both sides again: $$25x^2-200x+400=9(6x-17)$$ $$25x^2-254x+553=0$$ Solving this gives $$x=7\text{ or }x=\frac{79}{25}$$ but $x=\frac{79}{25}$ doesn't work in the original equation, so $x=7$ is the only solution to the original equation.
$8x+3=3x^2$
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
Best Answer
I am a bit surprised that nobody suggested $$\left({\frac{\sqrt 8}2}\right)^2 = \frac{{\left(\sqrt 8\right)}^2}{2^2} = \frac84 = 2. $$