10. Solve each radical expression. Check your solution.
1=√(-2v-3)
11. Solve each radical expression. Check your solution.
√(n+5)= √(5n-11)
12. Tell which solutions, if any, are extraneous for each question.
√(12-n)=n;
n= -4,n=3
#10. since 1=√1, you must have -2v-3 = 1
#11.
√(n+5)= √(5n-11)
n+5 = 5n-11
n = 4
check: √9 = √9
#12. well, geez, just check them!
√(12-(-4)) = √(12+4) = √16 = 4 Since this is not equal to -4, that value is extraneous.
√(12-3) = √9 = 3
yep, that one is ok.
To solve each of these radical expressions, we need to isolate the variable and then square both sides of the equation to eliminate the square root. Let's go through each problem step-by-step.
10. Solve the equation 1 = √(-2v - 3)
To isolate the square root, we need to move the constant term (-1) to the other side of the equation:
1 + √(-2v - 3) = 0
Next, we need to square both sides of the equation to eliminate the square root:
(1 + √(-2v - 3))^2 = 0^2
Simplifying the left side of the equation:
1 + 2√(-2v - 3) + (-2v - 3) = 0
Rearrange the equation:
2√(-2v - 3) - 2v - 2 = 0
Isolate the square root term:
2√(-2v - 3) = 2v + 2
Now, divide both sides of the equation by 2:
√(-2v - 3) = v + 1
Square both sides of the equation again to eliminate the square root:
√(-2v - 3)^2 = (v + 1)^2
-2v - 3 = v^2 + 2v + 1
Rearrange the equation:
v^2 + 4v + 4 = 0
Now, solve the quadratic equation by factoring or using the quadratic formula. However, upon solving this equation, we find that both solutions are extraneous. By substituting the values back into the original equation (1 = √(-2v - 3)), we can see that we end up with invalid solutions.
11. Solve the equation √(n + 5) = √(5n - 11)
Just like in the previous example, we need to isolate the square root term before squaring both sides of the equation:
√(n + 5)^2 = (√(5n - 11))^2
Simplifying:
n + 5 = 5n - 11
-4n = -16
Solve for n:
n = 4
Now, we can check if this solution is valid by substituting it back into the original equation (√(n + 5) = √(5n - 11)). Plugging in n = 4, we can see that both sides of the equation are equal:
√(4 + 5) = √(5*4 - 11)
√9 = √9
3 = 3
Hence, the solution n = 4 is valid.
12. Solve the equation √(12 - n) = n
To isolate the square root, we need to square both sides of the equation:
(√(12 - n))^2 = n^2
Simplifying:
12 - n = n^2
Rearranging the equation:
n^2 + n - 12 = 0
Now, we can solve the quadratic equation by factoring or using the quadratic formula:
(n - 3)(n + 4) = 0
From this equation, we find two solutions: n = 3 and n = -4.
To check for extraneous solutions, we substitute both values back into the original equation √(12 - n) = n:
For n = 3:
√(12 - 3) = 3
√9 = 3
3 = 3
For n = -4:
√(12 - (-4)) = -4
√16 = -4
4 = -4 (which is not true)
Hence, the solution n = -4 is extraneous and should be disregarded. Only n = 3 is a valid solution.