Basic Math Examples

$\sqrt{7 y^{2} + 27 y} = \sqrt{3 y^{2} + 40}$

Step 1

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

${\sqrt{7 y^{2} + 27 y}}^{2} = {\sqrt{3 y^{2} + 40}}^{2}$

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 $\sqrt{7 y^{2} + 27 y}$ as ${(7 y^{2} + 27 y)}^{\frac{1}{2}}$ .

${({(7 y^{2} + 27 y)}^{\frac{1}{2}})}^{2} = {\sqrt{3 y^{2} + 40}}^{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify ${({(7 y^{2} + 27 y)}^{\frac{1}{2}})}^{2}$ .

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Step 2.2.1.1

Multiply the exponents in ${({(7 y^{2} + 27 y)}^{\frac{1}{2}})}^{2}$ .

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Step 2.2.1.1.1

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

${(7 y^{2} + 27 y)}^{\frac{1}{2} \cdot 2} = {\sqrt{3 y^{2} + 40}}^{2}$

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.

${(7 y^{2} + 27 y)}^{\frac{1}{2} \cdot 2} = {\sqrt{3 y^{2} + 40}}^{2}$

Step 2.2.1.1.2.2

Rewrite the expression.

${(7 y^{2} + 27 y)}^{1} = {\sqrt{3 y^{2} + 40}}^{2}$

Step 2.2.1.2

Simplify.

$7 y^{2} + 27 y = {\sqrt{3 y^{2} + 40}}^{2}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Rewrite ${\sqrt{3 y^{2} + 40}}^{2}$ as $3 y^{2} + 40$ .

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Step 2.3.1.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 y^{2} + 40}$ as ${(3 y^{2} + 40)}^{\frac{1}{2}}$ .

$7 y^{2} + 27 y = {({(3 y^{2} + 40)}^{\frac{1}{2}})}^{2}$

Step 2.3.1.2

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

$7 y^{2} + 27 y = {(3 y^{2} + 40)}^{\frac{1}{2} \cdot 2}$

Step 2.3.1.3

Combine $\frac{1}{2}$ and $2$ .

$7 y^{2} + 27 y = {(3 y^{2} + 40)}^{\frac{2}{2}}$

Step 2.3.1.4

Cancel the common factor of $2$ .

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Step 2.3.1.4.1

Cancel the common factor.

$7 y^{2} + 27 y = {(3 y^{2} + 40)}^{\frac{2}{2}}$

Step 2.3.1.4.2

Rewrite the expression.

$7 y^{2} + 27 y = {(3 y^{2} + 40)}^{1}$

Step 2.3.1.5

Simplify.

$7 y^{2} + 27 y = 3 y^{2} + 40$

Step 3

Solve for $y$ .

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Step 3.1

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

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Step 3.1.1

Subtract $3 y^{2}$ from both sides of the equation.

$7 y^{2} + 27 y - 3 y^{2} = 40$

Step 3.1.2

Subtract $3 y^{2}$ from $7 y^{2}$ .

$4 y^{2} + 27 y = 40$

Step 3.2

Subtract $40$ from both sides of the equation.

$4 y^{2} + 27 y - 40 = 0$

Step 3.3

Factor by grouping.

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Step 3.3.1

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 $a \cdot c = 4 \cdot - 40 = - 160$ and whose sum is $b = 27$ .

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Step 3.3.1.1

Factor $27$ out of $27 y$ .

$4 y^{2} + 27 (y) - 40 = 0$

Step 3.3.1.2

Rewrite $27$ as $- 5$ plus $32$

$4 y^{2} + (- 5 + 32) y - 40 = 0$

Step 3.3.1.3

Apply the distributive property.

$4 y^{2} - 5 y + 32 y - 40 = 0$

Step 3.3.2

Factor out the greatest common factor from each group.

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Step 3.3.2.1

Group the first two terms and the last two terms.

$(4 y^{2} - 5 y) + 32 y - 40 = 0$

Step 3.3.2.2

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

$y (4 y - 5) + 8 (4 y - 5) = 0$

Step 3.3.3

Factor the polynomial by factoring out the greatest common factor, $4 y - 5$ .

$(4 y - 5) (y + 8) = 0$

Step 3.4

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

$4 y - 5 = 0$

$y + 8 = 0$

Step 3.5

Set $4 y - 5$ equal to $0$ and solve for $y$ .

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Step 3.5.1

Set $4 y - 5$ equal to $0$ .

$4 y - 5 = 0$

Step 3.5.2

Solve $4 y - 5 = 0$ for $y$ .

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Step 3.5.2.1

Add $5$ to both sides of the equation.

$4 y = 5$

Step 3.5.2.2

Divide each term in $4 y = 5$ by $4$ and simplify.

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Step 3.5.2.2.1

Divide each term in $4 y = 5$ by $4$ .

$\frac{4 y}{4} = \frac{5}{4}$

Step 3.5.2.2.2

Simplify the left side.

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Step 3.5.2.2.2.1

Cancel the common factor of $4$ .

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Step 3.5.2.2.2.1.1

Cancel the common factor.

$\frac{4 y}{4} = \frac{5}{4}$

Step 3.5.2.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{5}{4}$

Step 3.6

Set $y + 8$ equal to $0$ and solve for $y$ .

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Step 3.6.1

Set $y + 8$ equal to $0$ .

$y + 8 = 0$

Step 3.6.2

Subtract $8$ from both sides of the equation.

$y = - 8$

Step 3.7

The final solution is all the values that make $(4 y - 5) (y + 8) = 0$ true.

$y = \frac{5}{4}, - 8$

Step 4

The result can be shown in multiple forms.

Exact Form:

$y = \frac{5}{4}, - 8$

Decimal Form:

$y = 1.25, - 8$

Mixed Number Form:

$y = 1 \frac{1}{4}, - 8$