Basic Math Examples

$\sqrt{2 c - 1} = 2 c - 7$

Step 1

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{2 c - 1}}^{2} = {(2 c - 7)}^{2}$

Step 2

Simplify each side of the equation.

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Step 2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 c - 1}$ as ${(2 c - 1)}^{\frac{1}{2}}$ .

${({(2 c - 1)}^{\frac{1}{2}})}^{2} = {(2 c - 7)}^{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify ${({(2 c - 1)}^{\frac{1}{2}})}^{2}$ .

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Step 2.2.1.1

Multiply the exponents in ${({(2 c - 1)}^{\frac{1}{2}})}^{2}$ .

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Step 2.2.1.1.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(2 c - 1)}^{\frac{1}{2} \cdot 2} = {(2 c - 7)}^{2}$

Step 2.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.2.1.1.2.1

Cancel the common factor.

${(2 c - 1)}^{\frac{1}{2} \cdot 2} = {(2 c - 7)}^{2}$

Step 2.2.1.1.2.2

Rewrite the expression.

${(2 c - 1)}^{1} = {(2 c - 7)}^{2}$

Step 2.2.1.2

Simplify.

$2 c - 1 = {(2 c - 7)}^{2}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Simplify ${(2 c - 7)}^{2}$ .

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Step 2.3.1.1

Rewrite ${(2 c - 7)}^{2}$ as $(2 c - 7) (2 c - 7)$ .

$2 c - 1 = (2 c - 7) (2 c - 7)$

Step 2.3.1.2

Expand $(2 c - 7) (2 c - 7)$ using the FOIL Method.

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Step 2.3.1.2.1

Apply the distributive property.

$2 c - 1 = 2 c (2 c - 7) - 7 (2 c - 7)$

Step 2.3.1.2.2

Apply the distributive property.

$2 c - 1 = 2 c (2 c) + 2 c \cdot - 7 - 7 (2 c - 7)$

Step 2.3.1.2.3

Apply the distributive property.

$2 c - 1 = 2 c (2 c) + 2 c \cdot - 7 - 7 (2 c) - 7 \cdot - 7$

Step 2.3.1.3

Simplify and combine like terms.

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Step 2.3.1.3.1

Simplify each term.

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Step 2.3.1.3.1.1

Rewrite using the commutative property of multiplication.

$2 c - 1 = 2 \cdot 2 c \cdot c + 2 c \cdot - 7 - 7 (2 c) - 7 \cdot - 7$

Step 2.3.1.3.1.2

Multiply $c$ by $c$ by adding the exponents.

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Step 2.3.1.3.1.2.1

Move $c$ .

$2 c - 1 = 2 \cdot 2 (c \cdot c) + 2 c \cdot - 7 - 7 (2 c) - 7 \cdot - 7$

Step 2.3.1.3.1.2.2

Multiply $c$ by $c$ .

$2 c - 1 = 2 \cdot 2 c^{2} + 2 c \cdot - 7 - 7 (2 c) - 7 \cdot - 7$

Step 2.3.1.3.1.3

Multiply $2$ by $2$ .

$2 c - 1 = 4 c^{2} + 2 c \cdot - 7 - 7 (2 c) - 7 \cdot - 7$

Step 2.3.1.3.1.4

Multiply $- 7$ by $2$ .

$2 c - 1 = 4 c^{2} - 14 c - 7 (2 c) - 7 \cdot - 7$

Step 2.3.1.3.1.5

Multiply $2$ by $- 7$ .

$2 c - 1 = 4 c^{2} - 14 c - 14 c - 7 \cdot - 7$

Step 2.3.1.3.1.6

Multiply $- 7$ by $- 7$ .

$2 c - 1 = 4 c^{2} - 14 c - 14 c + 49$

Step 2.3.1.3.2

Subtract $14 c$ from $- 14 c$ .

$2 c - 1 = 4 c^{2} - 28 c + 49$

Step 3

Solve for $c$ .

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Step 3.1

Since $c$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$4 c^{2} - 28 c + 49 = 2 c - 1$

Step 3.2

Move all terms containing $c$ to the left side of the equation.

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Step 3.2.1

Subtract $2 c$ from both sides of the equation.

$4 c^{2} - 28 c + 49 - 2 c = - 1$

Step 3.2.2

Subtract $2 c$ from $- 28 c$ .

$4 c^{2} - 30 c + 49 = - 1$

Step 3.3

Add $1$ to both sides of the equation.

$4 c^{2} - 30 c + 49 + 1 = 0$

Step 3.4

Add $49$ and $1$ .

$4 c^{2} - 30 c + 50 = 0$

Step 3.5

Factor the left side of the equation.

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Step 3.5.1

Factor $2$ out of $4 c^{2} - 30 c + 50$ .

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Step 3.5.1.1

Factor $2$ out of $4 c^{2}$ .

$2 (2 c^{2}) - 30 c + 50 = 0$

Step 3.5.1.2

Factor $2$ out of $- 30 c$ .

$2 (2 c^{2}) + 2 (- 15 c) + 50 = 0$

Step 3.5.1.3

Factor $2$ out of $50$ .

$2 (2 c^{2}) + 2 (- 15 c) + 2 (25) = 0$

Step 3.5.1.4

Factor $2$ out of $2 (2 c^{2}) + 2 (- 15 c)$ .

$2 (2 c^{2} - 15 c) + 2 (25) = 0$

Step 3.5.1.5

Factor $2$ out of $2 (2 c^{2} - 15 c) + 2 (25)$ .

$2 (2 c^{2} - 15 c + 25) = 0$

Step 3.5.2

Factor.

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Step 3.5.2.1

Factor by grouping.

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Step 3.5.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 25 = 50$ and whose sum is $b = - 15$ .

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Step 3.5.2.1.1.1

Factor $- 15$ out of $- 15 c$ .

$2 (2 c^{2} - 15 c + 25) = 0$

Step 3.5.2.1.1.2

Rewrite $- 15$ as $- 5$ plus $- 10$

$2 (2 c^{2} + (- 5 - 10) c + 25) = 0$

Step 3.5.2.1.1.3

Apply the distributive property.

$2 (2 c^{2} - 5 c - 10 c + 25) = 0$

Step 3.5.2.1.2

Factor out the greatest common factor from each group.

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Step 3.5.2.1.2.1

Group the first two terms and the last two terms.

$2 ((2 c^{2} - 5 c) - 10 c + 25) = 0$

Step 3.5.2.1.2.2

Factor out the greatest common factor (GCF) from each group.

$2 (c (2 c - 5) - 5 (2 c - 5)) = 0$

Step 3.5.2.1.3

Factor the polynomial by factoring out the greatest common factor, $2 c - 5$ .

$2 ((2 c - 5) (c - 5)) = 0$

Step 3.5.2.2

Remove unnecessary parentheses.

$2 (2 c - 5) (c - 5) = 0$

Step 3.6

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 c - 5 = 0$

$c - 5 = 0$

Step 3.7

Set $2 c - 5$ equal to $0$ and solve for $c$ .

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Step 3.7.1

Set $2 c - 5$ equal to $0$ .

$2 c - 5 = 0$

Step 3.7.2

Solve $2 c - 5 = 0$ for $c$ .

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Step 3.7.2.1

Add $5$ to both sides of the equation.

$2 c = 5$

Step 3.7.2.2

Divide each term in $2 c = 5$ by $2$ and simplify.

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Step 3.7.2.2.1

Divide each term in $2 c = 5$ by $2$ .

$\frac{2 c}{2} = \frac{5}{2}$

Step 3.7.2.2.2

Simplify the left side.

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Step 3.7.2.2.2.1

Cancel the common factor of $2$ .

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Step 3.7.2.2.2.1.1

Cancel the common factor.

$\frac{2 c}{2} = \frac{5}{2}$

Step 3.7.2.2.2.1.2

Divide $c$ by $1$ .

$c = \frac{5}{2}$

Step 3.8

Set $c - 5$ equal to $0$ and solve for $c$ .

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Step 3.8.1

Set $c - 5$ equal to $0$ .

$c - 5 = 0$

Step 3.8.2

Add $5$ to both sides of the equation.

$c = 5$

Step 3.9

The final solution is all the values that make $2 (2 c - 5) (c - 5) = 0$ true.

$c = \frac{5}{2}, 5$

Step 4

Exclude the solutions that do not make $\sqrt{2 c - 1} = 2 c - 7$ true.

$c = 5$