Big Ideas Math Algebra 1, 2015
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Big Ideas Math Algebra 1, 2015 View details
3. Solving Radical Equations
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Exercise 69 Page 566

Practice makes perfect
a We are asked to solve the following radical equation by graphing.
2+x= sqrt(2x-3) To do so, we will first express each side of the equation as a function. Then, we will graph them on the same coordinate plane to see their point of intersection. The point of intersection will be the solution of the given equation.

g(x)=& 2+x f(x)=& sqrt(2x-3) To draw the graphs on a calculator, we first press the Y= button and type the functions in any of the rows. Having written the functions, we can push GRAPH to draw it.

As we can see, there is no point of intersection for these two graphs. Therefore, the given equation has no real solutions.

b This time we will solve the given radical equation algebraically. Let's begin by squaring both sides of the equation to eliminate the radical sign. Then we can solve for x.
(x+2)^() 2=(sqrt(2x-3))^2
x^2+4x+4=(sqrt(2x-3))^2
x^2+4x+4=2x-3
â–Ľ
Rewrite
x^2+4x+7=2x
x^2+2x+7=0
Now we have a quadratic equation. Therefore, we will continue solving by using the Quadratic Formula.
x^2+2x+7=0
x=- ( 2)±sqrt(( 2)^2-4( 1)( 7))/2( 1)
x=-2±sqrt(4-28)/2
x=-2±sqrt(-24)/2
We found that the value under the radical is -24. Since it is less than 0, the given equation has no real solutions.
c Let's think about the benefits of both solving methods and then you can decide which one works best for you!
Benefits of Solving Graphically Benefits of Solving Algebraically
We can immediately see whether or not there is a solution. When there is no solution, we do not need to continue any further. When there is a solution and we want to find its exact value, solving algebraically can be more accurate.

Sometimes, it can be helpful to try a combination of both methods. We can graph the equation as a pair of functions first to see whether or not there is a solution. Then, if there is a solution, we can solve algebraically to find its exact value.