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I've been stuck on this for a while. This is from Thinkwell P.8.

Here is the problem: $$(w+5)^\frac{1}{2}+6=5(w+5)^\frac{1}{4}$$

I need to find the solutions that allow the above to be true.

Before this class, it's been about 15 years since I've been in a math class, so I'm sure I'm missing something basic. Here's what I've been doing:

$$((w+5)^\frac{1}{2}+6)^4=(5(w+5)^\frac{1}{4})^4$$ $$(w+5)^2+6^4=5^4(w+5)$$ $$(w+5)(w+5)+1296=625(w+5)$$ $$w^2+10w+1321=625w+3125$$ $$w^2-615w+1321=3125$$ $$w^2-615w-1804=0$$

Now, feeding it through the quadratic formula leaves me with 2.91947426 and -925.4194743, neither of which turn into a clean fraction according to my TI-83.

I do know that the answer(s) should either be a whole number or a simple fraction.

Where am I going wrong?

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  • $\begingroup$ $(2+2)^2\ne2^2+2^2$ $\endgroup$
    – Simply Beautiful Art
    Commented Oct 19, 2016 at 20:40

3 Answers 3

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$((w+5)^\frac{1}{2}+6)^4=(5(w+5)^\frac{1}{4})^4$

$(w+5)^2+6^4=5^4(w+5)$

Whoa!!! $(a + b)^4 \ne a^4 + b^4$!!!!

$(w+5)^{\frac 12} + 6 = 5(w + 5)^{\frac 14}$. Let $(w+5)^{\frac 14} = \Omega$

$\Omega^2 + 6 = 5\Omega$

$\Omega^2 - 5\Omega + 6 = 0$

$(\Omega - 2)(\Omega - 3)=0$

$((w+5)^{\frac 14} - 2)((w+5)^{\frac 14} - 3) = 0$

So either

$(w+5)^{\frac 14} = 2$

or $(w+5)^{\frac 14} = 3$

So either $w + 5 = 16; w = 11$

or $w +5 = 81;w=76$

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  • $\begingroup$ Strange choice of variables. $\endgroup$
    – Simply Beautiful Art
    Commented Oct 19, 2016 at 21:35
  • $\begingroup$ This is much clearer to me, thank you. $\endgroup$
    – titanofold
    Commented Oct 19, 2016 at 21:42
  • $\begingroup$ Yes, it was a strange choice of variable. But I wanted to express (something that is trivial and irrelevant to experienced users but is oftern elusive to novices) that I wasn't actually substituting something to make a different equation-- the $\Omega$ were $(w+5)^{1/4} $ just written as $\Omega$ to prevent typing and eyestrain. $\endgroup$
    – fleablood
    Commented Oct 19, 2016 at 23:35
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Start with the substitution $x=(w+5)^{1/4}$

$$x^2+6=5x$$

$$x^2-5x+6=0$$

$$ \begin{align} \therefore 0 &= x^2-5x+6 \\ &= (x-2)(x-3) \\ \end{align} $$

So $x=2,3$, giving $w=11$ or $w=76$

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  • $\begingroup$ I see we posted the same response! $\endgroup$
    – Giuseppe
    Commented Oct 19, 2016 at 20:44
  • $\begingroup$ Unfortunately, neither of those are the solution. $\endgroup$
    – titanofold
    Commented Oct 19, 2016 at 20:59
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    $\begingroup$ Typo: $0=x^2 -5x +6$ so $0 = (x-2)(x-3)$ so $x=2,3$ so $w = 11,76$. $\endgroup$
    – fleablood
    Commented Oct 19, 2016 at 21:08
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First of all,

$(a+b)^n\neq a^n+b^n$, which you have written: $((w+5)^\frac{1}{2} + 6)^4\neq(w+5)^2+6^4$

Might I suggest letting $x=(w+5)^\frac{1}{4}$? then $x^2=(w+5)^\frac{1}{2}$ and it simplifies to a quadratic:

$x^2+6=5x$. Then once you solve for $x$, you need only compute $x^4=(w+5)$ for all values of $x$, then substitute the resulting values of $w$ into the original equation to make sure they're valid, i.e., if you're restricting $w$ to real numbers then $w+5\geq 0$.

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  • $\begingroup$ Okay, so (a+b)^n is not equal to a^n + b^n. How should the power be applied to the terms? $\endgroup$
    – titanofold
    Commented Oct 19, 2016 at 21:00
  • $\begingroup$ You don't. don't raise them to the fourth. Replace $(w+5)^{\frac 14} = x$ and $x^2 = (w+5)^{\frac 12}$. Then you have $x^2 + 6 = x$. Solve for x. Then you have $(w+5)^{\frac 14} = x$ so then you raise $w+5 = x^4$. $\endgroup$
    – fleablood
    Commented Oct 19, 2016 at 21:06
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    $\begingroup$ If you do want to multiply them out, see for example en.m.wikipedia.org/wiki/Binomial_theorem $\endgroup$
    – Giuseppe
    Commented Oct 19, 2016 at 21:10
  • $\begingroup$ However, I do not recommend this strategy for this particular problem $\endgroup$
    – Giuseppe
    Commented Oct 19, 2016 at 21:10

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