Question Find the solutions to the equation $(x-32)^{2}=56$ given $x= \pm 2 \sqrt{ } 22$ , $x= \pm 2 \sqrt{ } 6$ , $x=4 \sqrt{5} \pm 2 \sqrt{ } 14$ , and $x=32 \pm 2 \sqrt{ } 14$ .

Studdy Solution

STEP 1

Assumptions
1. The equation to solve is $(x-32)^{2}=56$ .
2. We are looking for real number solutions for $x$ .

STEP 2

Start by taking the square root of both sides of the equation to eliminate the exponent on the left-hand side.
$\sqrt{(x-32)^{2}} = \sqrt{56}$

STEP 3

Since the square root and the square cancel each other out on the left-hand side, we have:
$x - 32 = \pm\sqrt{56}$

STEP 4

Simplify the square root on the right-hand side by factoring 56 into its prime factors.
$56 = 2 \times 2 \times 2 \times 7$

STEP 5

Identify pairs of prime factors to take out of the square root.
$\sqrt{56} = \sqrt{2^2 \times 2 \times 7}$

STEP 6

Take the square root of the perfect square, which is $2^2$ , and leave the rest under the square root.
$\sqrt{56} = 2\sqrt{2 \times 7}$

STEP 7

Simplify the expression under the square root.
$\sqrt{56} = 2\sqrt{14}$

STEP 8

Substitute the simplified square root back into the equation.
$x - 32 = \pm 2\sqrt{14}$

STEP 9

Add 32 to both sides of the equation to solve for $x$ .
$x = 32 \pm 2\sqrt{14}$

STEP 10

Now we have the solutions for $x$ .
$x = 32 + 2\sqrt{14} \quad \text{or} \quad x = 32 - 2\sqrt{14}$
The solutions to the equation $(x-32)^{2}=56$ are $x = 32 + 2\sqrt{14}$ and $x = 32 - 2\sqrt{14}$ .

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Question Find the solutions to the equation $(x-32)^{2}=56$ given $x= \pm 2 \sqrt{ } 22$ , $x= \pm 2 \sqrt{ } 6$ , $x=4 \sqrt{5} \pm 2 \sqrt{ } 14$ , and $x=32 \pm 2 \sqrt{ } 14$ .

Studdy Solution

STEP 1

Assumptions
1. The equation to solve is $(x-32)^{2}=56$ .
2. We are looking for real number solutions for $x$ .

STEP 2

Start by taking the square root of both sides of the equation to eliminate the exponent on the left-hand side.
$\sqrt{(x-32)^{2}} = \sqrt{56}$

STEP 3

Since the square root and the square cancel each other out on the left-hand side, we have:
$x - 32 = \pm\sqrt{56}$

STEP 4

Simplify the square root on the right-hand side by factoring 56 into its prime factors.
$56 = 2 \times 2 \times 2 \times 7$

STEP 5

Identify pairs of prime factors to take out of the square root.
$\sqrt{56} = \sqrt{2^2 \times 2 \times 7}$

STEP 6

Take the square root of the perfect square, which is $2^2$ , and leave the rest under the square root.
$\sqrt{56} = 2\sqrt{2 \times 7}$

STEP 7

Simplify the expression under the square root.
$\sqrt{56} = 2\sqrt{14}$

STEP 8

Substitute the simplified square root back into the equation.
$x - 32 = \pm 2\sqrt{14}$

STEP 9

Add 32 to both sides of the equation to solve for $x$ .
$x = 32 \pm 2\sqrt{14}$

STEP 10

Now we have the solutions for $x$ .
$x = 32 + 2\sqrt{14} \quad \text{or} \quad x = 32 - 2\sqrt{14}$
The solutions to the equation $(x-32)^{2}=56$ are $x = 32 + 2\sqrt{14}$ and $x = 32 - 2\sqrt{14}$ .

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Question Find the solutions to the equation (x−32)2=56(x-32)^{2}=56(x−32)2=56 given x=±222x= \pm 2 \sqrt{ } 22x=±2​22, x=±26x= \pm 2 \sqrt{ } 6x=±2​6, x=45±214x=4 \sqrt{5} \pm 2 \sqrt{ } 14x=45​±2​14, and x=32±214x=32 \pm 2 \sqrt{ } 14x=32±2​14.

Studdy Solution

STEP 1

Assumptions1. The equation to solve is (x−32)2=56(x-32)^{2}=56(x−32)2=56.2. We are looking for real number solutions for xxx.

STEP 2

Start by taking the square root of both sides of the equation to eliminate the exponent on the left-hand side.(x−32)2=56\sqrt{(x-32)^{2}} = \sqrt{56}(x−32)2​=56​

STEP 3

Since the square root and the square cancel each other out on the left-hand side, we have:x−32=±56x - 32 = \pm\sqrt{56}x−32=±56​

STEP 4

Simplify the square root on the right-hand side by factoring 56 into its prime factors.56=2×2×2×756 = 2 \times 2 \times 2 \times 756=2×2×2×7

STEP 5

Identify pairs of prime factors to take out of the square root.56=22×2×7\sqrt{56} = \sqrt{2^2 \times 2 \times 7}56​=22×2×7​

STEP 6

Take the square root of the perfect square, which is 222^222, and leave the rest under the square root.56=22×7\sqrt{56} = 2\sqrt{2 \times 7}56​=22×7​

STEP 7

Simplify the expression under the square root.56=214\sqrt{56} = 2\sqrt{14}56​=214​

STEP 8

Substitute the simplified square root back into the equation.x−32=±214x - 32 = \pm 2\sqrt{14}x−32=±214​

STEP 9

Add 32 to both sides of the equation to solve for xxx.x=32±214x = 32 \pm 2\sqrt{14}x=32±214​

STEP 10

Now we have the solutions for xxx.x=32+214orx=32−214x = 32 + 2\sqrt{14} \quad \text{or} \quad x = 32 - 2\sqrt{14}x=32+214​orx=32−214​The solutions to the equation (x−32)2=56(x-32)^{2}=56(x−32)2=56 are x=32+214x = 32 + 2\sqrt{14}x=32+214​ and x=32−214x = 32 - 2\sqrt{14}x=32−214​.

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Question Find the solutions to the equation $(x-32)^{2}=56$ given $x= \pm 2 \sqrt{ } 22$ , $x= \pm 2 \sqrt{ } 6$ , $x=4 \sqrt{5} \pm 2 \sqrt{ } 14$ , and $x=32 \pm 2 \sqrt{ } 14$ .

Assumptions
1. The equation to solve is $(x-32)^{2}=56$ .
2. We are looking for real number solutions for $x$ .

Start by taking the square root of both sides of the equation to eliminate the exponent on the left-hand side.
$\sqrt{(x-32)^{2}} = \sqrt{56}$

Since the square root and the square cancel each other out on the left-hand side, we have:
$x - 32 = \pm\sqrt{56}$

Simplify the square root on the right-hand side by factoring 56 into its prime factors.
$56 = 2 \times 2 \times 2 \times 7$

Identify pairs of prime factors to take out of the square root.
$\sqrt{56} = \sqrt{2^2 \times 2 \times 7}$

Take the square root of the perfect square, which is $2^2$ , and leave the rest under the square root.
$\sqrt{56} = 2\sqrt{2 \times 7}$

Simplify the expression under the square root.
$\sqrt{56} = 2\sqrt{14}$

Substitute the simplified square root back into the equation.
$x - 32 = \pm 2\sqrt{14}$

Add 32 to both sides of the equation to solve for $x$ .
$x = 32 \pm 2\sqrt{14}$

Now we have the solutions for $x$ .
$x = 32 + 2\sqrt{14} \quad \text{or} \quad x = 32 - 2\sqrt{14}$
The solutions to the equation $(x-32)^{2}=56$ are $x = 32 + 2\sqrt{14}$ and $x = 32 - 2\sqrt{14}$ .