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Math  /  Algebra

QuestionSolve for vv, where vv is a real number. v=5v+24v=\sqrt{5 v+24}
If there is more than one solution, separate them with commas. If there is no solution, click on "No solution". v=v= □\square No solution
□\square

Studdy Solution

STEP 1

1. The equation v=5v+24 v = \sqrt{5v + 24} involves a square root.
2. We are looking for real solutions for v v .

STEP 2

1. Eliminate the square root by squaring both sides of the equation.
2. Rearrange the equation to form a standard quadratic equation.
3. Solve the quadratic equation for v v .
4. Verify the solutions to ensure they satisfy the original equation.

STEP 3

Square both sides of the equation to eliminate the square root:
v=5v+24 v = \sqrt{5v + 24}
Squaring both sides gives:
v2=5v+24 v^2 = 5v + 24

STEP 4

Rearrange the equation to form a standard quadratic equation:
v2=5v+24 v^2 = 5v + 24
Subtract 5v 5v and 24 24 from both sides:
v2−5v−24=0 v^2 - 5v - 24 = 0

STEP 5

Solve the quadratic equation v2−5v−24=0 v^2 - 5v - 24 = 0 using the quadratic formula:
The quadratic formula is:
v=−b±b2−4ac2a v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
For our equation, a=1 a = 1 , b=−5 b = -5 , and c=−24 c = -24 .
Calculate the discriminant:
b2−4ac=(−5)2−4×1×(−24) b^2 - 4ac = (-5)^2 - 4 \times 1 \times (-24) =25+96 = 25 + 96 =121 = 121
Since the discriminant is positive, there are two real solutions.
Calculate the solutions:
v=−(−5)±1212×1 v = \frac{-(-5) \pm \sqrt{121}}{2 \times 1} v=5±112 v = \frac{5 \pm 11}{2}
The two solutions are:
v=5+112=162=8 v = \frac{5 + 11}{2} = \frac{16}{2} = 8
v=5−112=−62=−3 v = \frac{5 - 11}{2} = \frac{-6}{2} = -3

STEP 6

Verify the solutions by substituting them back into the original equation:
For v=8 v = 8 :
8=5(8)+24 8 = \sqrt{5(8) + 24} 8=40+24 8 = \sqrt{40 + 24} 8=64 8 = \sqrt{64} 8=8 8 = 8 (True)
For v=−3 v = -3 :
−3=5(−3)+24 -3 = \sqrt{5(-3) + 24} −3=−15+24 -3 = \sqrt{-15 + 24} −3=9 -3 = \sqrt{9} −3=3 -3 = 3 (False)
The only valid solution is v=8 v = 8 .
The value of v v is:
8 \boxed{8}

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