Solving Radical Equations: Examples (page 2 of 6)

• Solve the equation:

The two lines represented by the two sides of this equation are:

...so you can see that there should be a solution at or about x = 10. To solve this algebraically, I need to square each side:

x – 1 = (x – 7)(x – 7)
x – 1 = x2 – 14x + 49

The squared expressions can be graphed as the lines y = x – 1 and y = x2 – 14x + 49. The solutions of x – 1 = x2 – 14x + 49 are the intersection points of the two lines:

As you can see, the intersection point at x = 10, from the first graph, is still there, but now a second, extraneous, solution has appeared at x = 5! ("Extraneous", pronounced as "ek-STRAY-nee-uss", in this context means "mathematically correct, but not relevant or useful, as far as the original question is concerned".) Continuing the solution:

x – 1 = x2 – 14x + 49
0 = x2 – 15x + 50

0 = (x – 5)(x – 10)

x = 5, x = 10

So I got the result that the second graph led me to expect, but I also know, from the first graph, that "x = 5" should not be a solution. This again illustrates why you always need to check your answers when solving radical equations: the very act of squaring has, in this case, produced an extra and incorrect "solution". Here's my check:

x = 5:

x = 10:

So the answer is x = 10.

• Solve the equation:

Since this equation is in the form "(square root) = (number)", I can proceed directly to squaring both sides:

x – 2 = 25
x = 27

This solution matches what I would expect from the graph of the two sides of the equation:

As you can see above, the lines:

y = 5

...intersect at x = 27, as the algebra had already shown me. Checking, I get:

So the solution is x = 27.

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