# SOLUTION: Solve the equation : 2(x)^1/2 = 9 + 5 (x)^-1/2 *Please answer as soon as possible. :) =)

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 Question 421189: Solve the equation : 2(x)^1/2 = 9 + 5 (x)^-1/2 *Please answer as soon as possible. :) =)Answer by jsmallt9(3296)   (Show Source): You can put this solution on YOUR website! When solving equations it is usually a good idea to start by simplifying each side of the equation. To me, part of simplifying is to eliminate fractions. (After all, aren't things "simpler" without fractions than with them?) There are two types of fractions here. There are the fractions in the exponents. And there is the "hidden" fraction represented by the negative exponent! This fraction will be obvious if we rewrite the equation with positive exponents: I'm going to eliminate the fraction. This can be done by multiplying both sides of the equation by : On the left side we use the rule for exponents for multiplying (i.e. add the exponents. On the right side we use the Distributive Property: which simplifies to: At this point we could eliminate the fraction that remains (in the exponent). Or we could recognize that the exponent of is twice the exponent of . This makes the equation an equation of "quadratic form". If you have trouble seeing this, use of temporary variable can help you see the "quadratic-ness" of the equation: Let Then Substituting these into the equation we get: I am going to solve this equation as a quadratic form equation. (At the end I will also solve the equation by eliminating the fraction in the exponent first.) Quadratic form equations are solved using the same techniques as "regular" quadratic equations. So we want one side to be zero. Since the rest of the problem will be a little easier if I keep the coefficient of the term with the biggest exponent positive, I am going to subtract the entire right side from both sides giving us: Now we factor (or use the Quadratic Formula). This factors farily easily: (2q+1)(q-5) = 0 From the Zero Product Property we know that one of these factors must be zero. So: 2q+1 = 0 or q-5 = 0 Solving these we get: q = -1/2 or q = 5 These are solutions for q. But we are interested in solutions for x. So we substitute back in for q. (Q was a temporary variable after all.) or Since an exponent of 1/2 means square root and since square roots cannot be negative, there is no solution to the first equation. To solve the second equation we just square both sides. This gives us: x = 25 Any time you square both sides of an equation you must check your answer(s). And when checking use, the original equation: Checking x = 25: which simplifies as follows: 10 = 9 + 1 10 = 10 Check! So x = 25 is the solution to your equation. If we don't notice that is an equation of quadratic form, or if we just decide we don't like solving quadratic form equations, we can solve this equation as a square root equation (since an exponent of 1/2 means square root. To start I am going to actuall write the equation with a square root: A procedure for solving square root equations is:Isolate a square root term that has a variable in its radicand. (The expression inside a radical is called a radicand.)Square both sides of the equation.If there is still a square root with a variable in its radicand, then repeat steps 1 and 2.At this point there should be no square roots with a variable in its radicand. Solve the equation using techniques appropriate for the type of equation it is.Since you have squared both sides of the equation at least once at step 2, you must check your solutions.Let's see this in action: 1) Isolate a square root. Subtracting 5 from each side we get: (We can leave the 9 where it is or we can divide both sides of the equation by 9. The important thing is: will the sides of the equation be easier to square with the 9 where it is or with the 9 in the denominator on the other side? I think it is easier where it is. If we had something like I would multiply both sides by 2.) 2. Square both sides: The right side is easy to square. On the left side we should use FOIL or the pattern. I like using the pattern: which simplifies to: 3) If there is still a square root... The square root is gone. 4) Solve the equation. This is a quadratic equation. So we want one side equal to zero. Subtracting 81x from each side we get: Now we factor (or use the Quadratic Formula). This does factor: (4x-1)(x-25) = 0 From the Zero Product Property we knwo that one of these factors must be zero. So: 4x-1 = 0 or x-25 = 0 Solving these we get: x = 1/4 or x = 25 5) Check you answer(s). Usually you go back to the original equation. And we can do so here. But any of the equations we had before we squared both sides can be used. I'm going to use: Checking x = 1/4: which simplifies as follows: Check fails!! So we reject x = 1/4 as a solution. (x = 1/4 is what is called an extraneous solution. Extraneous solutions are solutions that work in the squared equation but not in the "pre-squared" equation. Extraneous solutions can happen any time you square both sides of an equation. Extraneous solutions may happen even if no mistakes have been made. This is why even expert mathematicians must check their answers after they have squared both sides of an equation.) Checking x = 25: 50 = 9*5 + 5 50 = 45 + 5 50 = 50 Check!! So the only solution, just as found with the quadratic form solution, is x = 25.