What Integral Values Satisfy This Radical Equation?

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    2015
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SUMMARY

The radical equation $\sqrt{a+8-6\sqrt{a-1}}+\sqrt{a+3-4\sqrt{a-1}}=1$ is solved for integral values of $a$. The correct solutions identified by members include $a = 1$ and $a = 4$. The discussion highlights the importance of manipulating radical expressions and understanding the properties of square roots in solving such equations.

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anemone
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Here is this week's POTW:

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Find all integral values of $a$ such that $\sqrt{a+8-6\sqrt{a-1}}+\sqrt{a+3-4\sqrt{a-1}}=1$.

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Congratulations to the following members for their correct solution::)

1. MarkFL
2. kaliprasad
3. lfdahl
4. greg1313

Solution from MarkFL:
Let:

$$a-1=m^2\implies a=m^2+1$$

And we have:

$$\sqrt{m^2-6m+9}+\sqrt{m^2-4m+4}=1$$

Or:

$$|m-3|+|m-2|=1$$

We have the following 3 cases to consider:

i) $$3\le m$$

$$m-3+m-2=1$$

$$m=3$$

ii) $$2<m<3$$

$$3-m+m-2=1$$

$$1=1$$

Because this is an identity, we know all $m$ in this interval is a solution.

iii) $$m\le2$$

$$3-m+2-m=1$$

$$m=2$$

Hence we find:

$$2\le m\le 3$$

Square through, and since all are positive, the inequality does not change direction:

$$4\le m^2\le9$$

Add through by 1:

$$5\le m^2+1\le10$$

Hence:

$$4\le a\le10$$

Since $a$ is integral, we conclude:

$$a\in\{5,6,7,8,9,10\}$$
 

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