Math / Algebra

QuestionSolving Radical Equations Solve each of the following equations algebraically. Check your answer graphically. Write you answer as an equation using Integer Values or Reduced Fractions.

Studdy Solution

STEP 1

1. We are solving radical equations of the form $a\sqrt{bx + c} + d = e$ or similar.
2. Solutions should be checked graphically to ensure they are valid within the domain of the original equation.
3. Solutions should be presented as integer values or reduced fractions.

STEP 2

1. Isolate the radical on one side of the equation.
2. Square both sides to eliminate the radical.
3. Solve the resulting algebraic equation.
4. Check for extraneous solutions by substituting back into the original equation.
5. Verify solutions graphically if required.

STEP 3

Isolate the radical in the first equation: $2\sqrt{x} + 6 = 8$ .
$2\sqrt{x} + 6 = 8$ $2\sqrt{x} = 8 - 6$ $2\sqrt{x} = 2$ $\sqrt{x} = 1$

STEP 4

Square both sides to eliminate the radical.
$(\sqrt{x})^2 = 1^2$ $x = 1$

STEP 5

Check the solution by substituting $x = 1$ back into the original equation.
$2\sqrt{1} + 6 = 8$ $2 \cdot 1 + 6 = 8$ $8 = 8$ Thus, $x = 1$ is correct.

STEP 6

Isolate the radical in the second equation: $\sqrt{7x + 8} - 5 = 0$ .
$\sqrt{7x + 8} - 5 = 0$ $\sqrt{7x + 8} = 5$

STEP 7

Square both sides to eliminate the radical.
$(\sqrt{7x + 8})^2 = 5^2$ $7x + 8 = 25$

STEP 8

Solve the resulting algebraic equation.
$7x + 8 = 25$ $7x = 25 - 8$ $7x = 17$ $x = \frac{17}{7}$

STEP 9

Check the solution by substituting $x = \frac{17}{7}$ back into the original equation.
$\sqrt{7 \left( \frac{17}{7} \right) + 8} - 5 = 0$ $\sqrt{17 + 8} - 5 = 0$ $\sqrt{25} - 5 = 0$ $5 - 5 = 0$ Thus, $x = \frac{17}{7}$ is correct.

STEP 10

Isolate the radical in the third equation: $2\sqrt{3x} - 6 = 8$ .
$2\sqrt{3x} - 6 = 8$ $2\sqrt{3x} = 8 + 6$ $2\sqrt{3x} = 14$ $\sqrt{3x} = 7$

STEP 11

Square both sides to eliminate the radical.
$(\sqrt{3x})^2 = 7^2$ $3x = 49$

STEP 12

Solve the resulting algebraic equation.
$3x = 49$ $x = \frac{49}{3}$

STEP 13

Check the solution by substituting $x = \frac{49}{3}$ back into the original equation.
$2\sqrt{3 \left( \frac{49}{3} \right)} - 6 = 8$ $2\sqrt{49} - 6 = 8$ $2 \cdot 7 - 6 = 8$ $14 - 6 = 8$ $8 = 8$ Thus, $x = \frac{49}{3}$ is correct.

STEP 14

Isolate the radicals in the fourth equation: $\sqrt{5x + 8} = \sqrt{5x + 15}$ .
$\sqrt{5x + 8} = \sqrt{5x + 15}$

STEP 15

Square both sides to eliminate the radicals.
$(\sqrt{5x + 8})^2 = (\sqrt{5x + 15})^2$ $5x + 8 = 5x + 15$

STEP 16

Solve the resulting algebraic equation.
$5x + 8 = 5x + 15$ Subtract $5x$ from both sides: $8 = 15$ This is a contradiction, meaning there are no solutions.
Thus, for the fourth equation, there are no valid solutions.

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QuestionSolving Radical Equations Solve each of the following equations algebraically. Check your answer graphically. Write you answer as an equation using Integer Values or Reduced Fractions.

Studdy Solution

STEP 1

1. We are solving radical equations of the form $a\sqrt{bx + c} + d = e$ or similar.
2. Solutions should be checked graphically to ensure they are valid within the domain of the original equation.
3. Solutions should be presented as integer values or reduced fractions.

STEP 2

1. Isolate the radical on one side of the equation.
2. Square both sides to eliminate the radical.
3. Solve the resulting algebraic equation.
4. Check for extraneous solutions by substituting back into the original equation.
5. Verify solutions graphically if required.

STEP 3

Isolate the radical in the first equation: $2\sqrt{x} + 6 = 8$ .
$2\sqrt{x} + 6 = 8$ $2\sqrt{x} = 8 - 6$ $2\sqrt{x} = 2$ $\sqrt{x} = 1$

STEP 4

Square both sides to eliminate the radical.
$(\sqrt{x})^2 = 1^2$ $x = 1$

STEP 5

Check the solution by substituting $x = 1$ back into the original equation.
$2\sqrt{1} + 6 = 8$ $2 \cdot 1 + 6 = 8$ $8 = 8$ Thus, $x = 1$ is correct.

STEP 6

Isolate the radical in the second equation: $\sqrt{7x + 8} - 5 = 0$ .
$\sqrt{7x + 8} - 5 = 0$ $\sqrt{7x + 8} = 5$

STEP 7

Square both sides to eliminate the radical.
$(\sqrt{7x + 8})^2 = 5^2$ $7x + 8 = 25$

STEP 8

Solve the resulting algebraic equation.
$7x + 8 = 25$ $7x = 25 - 8$ $7x = 17$ $x = \frac{17}{7}$

STEP 9

Check the solution by substituting $x = \frac{17}{7}$ back into the original equation.
$\sqrt{7 \left( \frac{17}{7} \right) + 8} - 5 = 0$ $\sqrt{17 + 8} - 5 = 0$ $\sqrt{25} - 5 = 0$ $5 - 5 = 0$ Thus, $x = \frac{17}{7}$ is correct.

STEP 10

Isolate the radical in the third equation: $2\sqrt{3x} - 6 = 8$ .
$2\sqrt{3x} - 6 = 8$ $2\sqrt{3x} = 8 + 6$ $2\sqrt{3x} = 14$ $\sqrt{3x} = 7$

STEP 11

Square both sides to eliminate the radical.
$(\sqrt{3x})^2 = 7^2$ $3x = 49$

STEP 12

Solve the resulting algebraic equation.
$3x = 49$ $x = \frac{49}{3}$

STEP 13

Check the solution by substituting $x = \frac{49}{3}$ back into the original equation.
$2\sqrt{3 \left( \frac{49}{3} \right)} - 6 = 8$ $2\sqrt{49} - 6 = 8$ $2 \cdot 7 - 6 = 8$ $14 - 6 = 8$ $8 = 8$ Thus, $x = \frac{49}{3}$ is correct.

STEP 14

Isolate the radicals in the fourth equation: $\sqrt{5x + 8} = \sqrt{5x + 15}$ .
$\sqrt{5x + 8} = \sqrt{5x + 15}$

STEP 15

Square both sides to eliminate the radicals.
$(\sqrt{5x + 8})^2 = (\sqrt{5x + 15})^2$ $5x + 8 = 5x + 15$

STEP 16

Solve the resulting algebraic equation.
$5x + 8 = 5x + 15$ Subtract $5x$ from both sides: $8 = 15$ This is a contradiction, meaning there are no solutions.
Thus, for the fourth equation, there are no valid solutions.

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QuestionSolving Radical Equations Solve each of the following equations algebraically. Check your answer graphically. Write you answer as an equation using Integer Values or Reduced Fractions.

Studdy Solution

STEP 1

1. We are solving radical equations of the form abx+c+d=e a\sqrt{bx + c} + d = e abx+c​+d=e or similar.2. Solutions should be checked graphically to ensure they are valid within the domain of the original equation.3. Solutions should be presented as integer values or reduced fractions.

STEP 2

1. Isolate the radical on one side of the equation.2. Square both sides to eliminate the radical.3. Solve the resulting algebraic equation.4. Check for extraneous solutions by substituting back into the original equation.5. Verify solutions graphically if required.

STEP 3

Isolate the radical in the first equation: 2x+6=8 2\sqrt{x} + 6 = 8 2x​+6=8.2x+6=8 2\sqrt{x} + 6 = 8 2x​+6=8 2x=8−6 2\sqrt{x} = 8 - 6 2x​=8−6 2x=2 2\sqrt{x} = 2 2x​=2 x=1 \sqrt{x} = 1 x​=1

STEP 4

Square both sides to eliminate the radical.(x)2=12 (\sqrt{x})^2 = 1^2 (x​)2=12 x=1 x = 1 x=1

STEP 5

Check the solution by substituting x=1 x = 1 x=1 back into the original equation.21+6=8 2\sqrt{1} + 6 = 8 21​+6=8 2⋅1+6=8 2 \cdot 1 + 6 = 8 2⋅1+6=8 8=8 8 = 8 8=8 Thus, x=1 x = 1 x=1 is correct.

STEP 6

Isolate the radical in the second equation: 7x+8−5=0 \sqrt{7x + 8} - 5 = 0 7x+8​−5=0.7x+8−5=0 \sqrt{7x + 8} - 5 = 0 7x+8​−5=0 7x+8=5 \sqrt{7x + 8} = 5 7x+8​=5

STEP 7

Square both sides to eliminate the radical.(7x+8)2=52 (\sqrt{7x + 8})^2 = 5^2 (7x+8​)2=52 7x+8=25 7x + 8 = 25 7x+8=25

STEP 8

Solve the resulting algebraic equation.7x+8=25 7x + 8 = 25 7x+8=25 7x=25−8 7x = 25 - 8 7x=25−8 7x=17 7x = 17 7x=17 x=177 x = \frac{17}{7} x=717​

STEP 9

STEP 10

Isolate the radical in the third equation: 23x−6=8 2\sqrt{3x} - 6 = 8 23x​−6=8.23x−6=8 2\sqrt{3x} - 6 = 8 23x​−6=8 23x=8+6 2\sqrt{3x} = 8 + 6 23x​=8+6 23x=14 2\sqrt{3x} = 14 23x​=14 3x=7 \sqrt{3x} = 7 3x​=7

STEP 11

Square both sides to eliminate the radical.(3x)2=72 (\sqrt{3x})^2 = 7^2 (3x​)2=72 3x=49 3x = 49 3x=49

STEP 12

Solve the resulting algebraic equation.3x=49 3x = 49 3x=49 x=493 x = \frac{49}{3} x=349​

STEP 13

STEP 14

Isolate the radicals in the fourth equation: 5x+8=5x+15 \sqrt{5x + 8} = \sqrt{5x + 15} 5x+8​=5x+15​.5x+8=5x+15 \sqrt{5x + 8} = \sqrt{5x + 15} 5x+8​=5x+15​

STEP 15

Square both sides to eliminate the radicals.(5x+8)2=(5x+15)2 (\sqrt{5x + 8})^2 = (\sqrt{5x + 15})^2 (5x+8​)2=(5x+15​)2 5x+8=5x+15 5x + 8 = 5x + 15 5x+8=5x+15

STEP 16

Solve the resulting algebraic equation.5x+8=5x+15 5x + 8 = 5x + 15 5x+8=5x+15 Subtract 5x 5x 5x from both sides: 8=15 8 = 15 8=15 This is a contradiction, meaning there are no solutions.Thus, for the fourth equation, there are no valid solutions.

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1. We are solving radical equations of the form $a\sqrt{bx + c} + d = e$ or similar.
2. Solutions should be checked graphically to ensure they are valid within the domain of the original equation.
3. Solutions should be presented as integer values or reduced fractions.

1. Isolate the radical on one side of the equation.
2. Square both sides to eliminate the radical.
3. Solve the resulting algebraic equation.
4. Check for extraneous solutions by substituting back into the original equation.
5. Verify solutions graphically if required.

Isolate the radical in the first equation: $2\sqrt{x} + 6 = 8$ .
$2\sqrt{x} + 6 = 8$ $2\sqrt{x} = 8 - 6$ $2\sqrt{x} = 2$ $\sqrt{x} = 1$

Square both sides to eliminate the radical.
$(\sqrt{x})^2 = 1^2$ $x = 1$

Check the solution by substituting $x = 1$ back into the original equation.
$2\sqrt{1} + 6 = 8$ $2 \cdot 1 + 6 = 8$ $8 = 8$ Thus, $x = 1$ is correct.

Isolate the radical in the second equation: $\sqrt{7x + 8} - 5 = 0$ .
$\sqrt{7x + 8} - 5 = 0$ $\sqrt{7x + 8} = 5$

Square both sides to eliminate the radical.
$(\sqrt{7x + 8})^2 = 5^2$ $7x + 8 = 25$

Solve the resulting algebraic equation.
$7x + 8 = 25$ $7x = 25 - 8$ $7x = 17$ $x = \frac{17}{7}$

Isolate the radical in the third equation: $2\sqrt{3x} - 6 = 8$ .
$2\sqrt{3x} - 6 = 8$ $2\sqrt{3x} = 8 + 6$ $2\sqrt{3x} = 14$ $\sqrt{3x} = 7$

Square both sides to eliminate the radical.
$(\sqrt{3x})^2 = 7^2$ $3x = 49$

Solve the resulting algebraic equation.
$3x = 49$ $x = \frac{49}{3}$

Isolate the radicals in the fourth equation: $\sqrt{5x + 8} = \sqrt{5x + 15}$ .
$\sqrt{5x + 8} = \sqrt{5x + 15}$

Square both sides to eliminate the radicals.
$(\sqrt{5x + 8})^2 = (\sqrt{5x + 15})^2$ $5x + 8 = 5x + 15$

Solve the resulting algebraic equation.
$5x + 8 = 5x + 15$ Subtract $5x$ from both sides: $8 = 15$ This is a contradiction, meaning there are no solutions.
Thus, for the fourth equation, there are no valid solutions.