Solved on Sep 18, 2023

Find the value of $v$ that satisfies the equation $v = \sqrt{5v + 6}$ . If there are multiple solutions, list them. If no solution exists, select "No solution".

STEP 1

Assumptions1. We are given the equation $v=\sqrt{5v+6}$ . . We are asked to solve for $v$ .

STEP 2

To solve for $v$ , we first square both sides of the equation to eliminate the square root. This gives us $v^2 = (5v+6)$

STEP 3

Rearrange the equation to set it equal to zero. This will allow us to use the quadratic formula to solve for $v$ .
$v^2 -5v -6 =0$

STEP 4

The quadratic formula is given by $v = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$ where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c =0$ .

STEP 5

In our equation, $a=1$ , $b=-5$ , and $c=-$ . Substituting these values into the quadratic formula gives us $v = \frac{-(-5) \pm \sqrt{(-5)^2 -4(1)(-)}}{2(1)}$

STEP 6

implify the equation to get $v = \frac{5 \pm \sqrt{25 +24}}{2}$

STEP 7

Further simplify the equation to get $v = \frac{5 \pm \sqrt{49}}{2}$

STEP 8

implify the square root to get $v = \frac{5 \pm7}{2}$

STEP 9

This gives us two possible solutions for $v$ $v = \frac{5 +7}{2} =6$ and $v = \frac{5 -7}{2} = -$

STEP 10

However, we must check these solutions in the original equation to make sure they are valid. Substituting $v=6$ into the original equation gives us $6 = \sqrt{5(6) +6} = \sqrt{36} =6$ So $v=6$ is a valid solution.

STEP 11

Substituting $v=-$ into the original equation gives us $- = \sqrt{5(-) +6} = \sqrt{} =$ This is not true, so $v=-$ is not a valid solution.
The only solution to the equation $v=\sqrt{5v+6}$ is $v=6$ .

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Find the value of $v$ that satisfies the equation $v = \sqrt{5v + 6}$ . If there are multiple solutions, list them. If no solution exists, select "No solution".

STEP 1

Assumptions1. We are given the equation $v=\sqrt{5v+6}$ . . We are asked to solve for $v$ .

STEP 2

To solve for $v$ , we first square both sides of the equation to eliminate the square root. This gives us $v^2 = (5v+6)$

STEP 3

Rearrange the equation to set it equal to zero. This will allow us to use the quadratic formula to solve for $v$ .
$v^2 -5v -6 =0$

STEP 4

The quadratic formula is given by $v = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$ where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c =0$ .

STEP 5

In our equation, $a=1$ , $b=-5$ , and $c=-$ . Substituting these values into the quadratic formula gives us $v = \frac{-(-5) \pm \sqrt{(-5)^2 -4(1)(-)}}{2(1)}$

STEP 6

implify the equation to get $v = \frac{5 \pm \sqrt{25 +24}}{2}$

STEP 7

Further simplify the equation to get $v = \frac{5 \pm \sqrt{49}}{2}$

STEP 8

implify the square root to get $v = \frac{5 \pm7}{2}$

STEP 9

This gives us two possible solutions for $v$ $v = \frac{5 +7}{2} =6$ and $v = \frac{5 -7}{2} = -$

STEP 10

However, we must check these solutions in the original equation to make sure they are valid. Substituting $v=6$ into the original equation gives us $6 = \sqrt{5(6) +6} = \sqrt{36} =6$ So $v=6$ is a valid solution.

STEP 11

Substituting $v=-$ into the original equation gives us $- = \sqrt{5(-) +6} = \sqrt{} =$ This is not true, so $v=-$ is not a valid solution.
The only solution to the equation $v=\sqrt{5v+6}$ is $v=6$ .

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Find the value of vvv that satisfies the equation v=5v+6v = \sqrt{5v + 6}v=5v+6​. If there are multiple solutions, list them. If no solution exists, select "No solution".

STEP 1

Assumptions1. We are given the equation v=5v+6v=\sqrt{5v+6}v=5v+6​. . We are asked to solve for vvv.

STEP 2

To solve for vvv, we first square both sides of the equation to eliminate the square root. This gives usv2=(5v+6)v^2 = (5v+6)v2=(5v+6)

STEP 3

Rearrange the equation to set it equal to zero. This will allow us to use the quadratic formula to solve for vvv.v2−5v−6=0v^2 -5v -6 =0v2−5v−6=0

STEP 4

The quadratic formula is given byv=−b±b2−4ac2av = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}v=2a−b±b2−4ac​​where aaa, bbb, and ccc are the coefficients of the quadratic equation ax2+bx+c=0ax^2 + bx + c =0ax2+bx+c=0.

STEP 5

In our equation, a=1a=1a=1, b=−5b=-5b=−5, and c=−c=-c=−. Substituting these values into the quadratic formula gives usv=−(−5)±(−5)2−4(1)(−)2(1)v = \frac{-(-5) \pm \sqrt{(-5)^2 -4(1)(-)}}{2(1)}v=2(1)−(−5)±(−5)2−4(1)(−)​​

STEP 6

implify the equation to getv=5±25+242v = \frac{5 \pm \sqrt{25 +24}}{2}v=25±25+24​​

STEP 7

Further simplify the equation to getv=5±492v = \frac{5 \pm \sqrt{49}}{2}v=25±49​​

STEP 8

implify the square root to getv=5±72v = \frac{5 \pm7}{2}v=25±7​

STEP 9

This gives us two possible solutions for vvvv=5+72=6v = \frac{5 +7}{2} =6v=25+7​=6andv=5−72=−v = \frac{5 -7}{2} = -v=25−7​=−

STEP 10

However, we must check these solutions in the original equation to make sure they are valid. Substituting v=6v=6v=6 into the original equation gives us6=5(6)+6=36=66 = \sqrt{5(6) +6} = \sqrt{36} =66=5(6)+6​=36​=6So v=6v=6v=6 is a valid solution.

STEP 11

Substituting v=−v=-v=− into the original equation gives us−=5(−)+6==- = \sqrt{5(-) +6} = \sqrt{} =−=5(−)+6​=​=This is not true, so v=−v=-v=− is not a valid solution.The only solution to the equation v=5v+6v=\sqrt{5v+6}v=5v+6​ is v=6v=6v=6.

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Find the value of $v$ that satisfies the equation $v = \sqrt{5v + 6}$ . If there are multiple solutions, list them. If no solution exists, select "No solution".

Assumptions1. We are given the equation $v=\sqrt{5v+6}$ . . We are asked to solve for $v$ .

To solve for $v$ , we first square both sides of the equation to eliminate the square root. This gives us $v^2 = (5v+6)$

Rearrange the equation to set it equal to zero. This will allow us to use the quadratic formula to solve for $v$ .
$v^2 -5v -6 =0$

The quadratic formula is given by $v = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$ where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c =0$ .

In our equation, $a=1$ , $b=-5$ , and $c=-$ . Substituting these values into the quadratic formula gives us $v = \frac{-(-5) \pm \sqrt{(-5)^2 -4(1)(-)}}{2(1)}$

implify the equation to get $v = \frac{5 \pm \sqrt{25 +24}}{2}$

Further simplify the equation to get $v = \frac{5 \pm \sqrt{49}}{2}$

implify the square root to get $v = \frac{5 \pm7}{2}$

This gives us two possible solutions for $v$ $v = \frac{5 +7}{2} =6$ and $v = \frac{5 -7}{2} = -$

However, we must check these solutions in the original equation to make sure they are valid. Substituting $v=6$ into the original equation gives us $6 = \sqrt{5(6) +6} = \sqrt{36} =6$ So $v=6$ is a valid solution.

Substituting $v=-$ into the original equation gives us $- = \sqrt{5(-) +6} = \sqrt{} =$ This is not true, so $v=-$ is not a valid solution.
The only solution to the equation $v=\sqrt{5v+6}$ is $v=6$ .