Precalculus Examples

$2 y - 3 = \sqrt{3 y^{2} - 10 y + 12}$

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

$\sqrt{3 y^{2} - 10 y + 12} = 2 y - 3$

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

${\sqrt{3 y^{2} - 10 y + 12}}^{2} = {(2 y - 3)}^{2}$

Simplify each side of the equation.

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 y^{2} - 10 y + 12}$ as ${(3 y^{2} - 10 y + 12)}^{\frac{1}{2}}$ .

${({(3 y^{2} - 10 y + 12)}^{\frac{1}{2}})}^{2} = {(2 y - 3)}^{2}$

Simplify the left side.

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Simplify ${({(3 y^{2} - 10 y + 12)}^{\frac{1}{2}})}^{2}$ .

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Multiply the exponents in ${({(3 y^{2} - 10 y + 12)}^{\frac{1}{2}})}^{2}$ .

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

${(3 y^{2} - 10 y + 12)}^{\frac{1}{2} \cdot 2} = {(2 y - 3)}^{2}$

Cancel the common factor of $2$ .

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Cancel the common factor.

${(3 y^{2} - 10 y + 12)}^{\frac{1}{2} \cdot 2} = {(2 y - 3)}^{2}$

Rewrite the expression.

${(3 y^{2} - 10 y + 12)}^{1} = {(2 y - 3)}^{2}$

Simplify.

$3 y^{2} - 10 y + 12 = {(2 y - 3)}^{2}$

Simplify the right side.

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Simplify ${(2 y - 3)}^{2}$ .

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Rewrite ${(2 y - 3)}^{2}$ as $(2 y - 3) (2 y - 3)$ .

$3 y^{2} - 10 y + 12 = (2 y - 3) (2 y - 3)$

Expand $(2 y - 3) (2 y - 3)$ using the FOIL Method.

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Apply the distributive property.

$3 y^{2} - 10 y + 12 = 2 y (2 y - 3) - 3 (2 y - 3)$

Apply the distributive property.

$3 y^{2} - 10 y + 12 = 2 y (2 y) + 2 y \cdot - 3 - 3 (2 y - 3)$

Apply the distributive property.

$3 y^{2} - 10 y + 12 = 2 y (2 y) + 2 y \cdot - 3 - 3 (2 y) - 3 \cdot - 3$

Simplify and combine like terms.

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Simplify each term.

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Rewrite using the commutative property of multiplication.

$3 y^{2} - 10 y + 12 = 2 \cdot 2 y \cdot y + 2 y \cdot - 3 - 3 (2 y) - 3 \cdot - 3$

Multiply $y$ by $y$ by adding the exponents.

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Move $y$ .

$3 y^{2} - 10 y + 12 = 2 \cdot 2 (y \cdot y) + 2 y \cdot - 3 - 3 (2 y) - 3 \cdot - 3$

Multiply $y$ by $y$ .

$3 y^{2} - 10 y + 12 = 2 \cdot 2 y^{2} + 2 y \cdot - 3 - 3 (2 y) - 3 \cdot - 3$

Multiply $2$ by $2$ .

$3 y^{2} - 10 y + 12 = 4 y^{2} + 2 y \cdot - 3 - 3 (2 y) - 3 \cdot - 3$

Multiply $- 3$ by $2$ .

$3 y^{2} - 10 y + 12 = 4 y^{2} - 6 y - 3 (2 y) - 3 \cdot - 3$

Multiply $2$ by $- 3$ .

$3 y^{2} - 10 y + 12 = 4 y^{2} - 6 y - 6 y - 3 \cdot - 3$

Multiply $- 3$ by $- 3$ .

$3 y^{2} - 10 y + 12 = 4 y^{2} - 6 y - 6 y + 9$

Subtract $6 y$ from $- 6 y$ .

$3 y^{2} - 10 y + 12 = 4 y^{2} - 12 y + 9$

Solve for $y$ .

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

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Subtract $4 y^{2}$ from both sides of the equation.

$3 y^{2} - 10 y + 12 - 4 y^{2} = - 12 y + 9$

Add $12 y$ to both sides of the equation.

$3 y^{2} - 10 y + 12 - 4 y^{2} + 12 y = 9$

Subtract $4 y^{2}$ from $3 y^{2}$ .

$- y^{2} - 10 y + 12 + 12 y = 9$

Add $- 10 y$ and $12 y$ .

$- y^{2} + 2 y + 12 = 9$

Subtract $9$ from both sides of the equation.

$- y^{2} + 2 y + 12 - 9 = 0$

Subtract $9$ from $12$ .

$- y^{2} + 2 y + 3 = 0$

Factor the left side of the equation.

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Factor $- 1$ out of $- y^{2} + 2 y + 3$ .

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Factor $- 1$ out of $- y^{2}$ .

$- (y^{2}) + 2 y + 3 = 0$

Factor $- 1$ out of $2 y$ .

$- (y^{2}) - (- 2 y) + 3 = 0$

Rewrite $3$ as $- 1 (- 3)$ .

$- (y^{2}) - (- 2 y) - 1 \cdot - 3 = 0$

Factor $- 1$ out of $- (y^{2}) - (- 2 y)$ .

$- (y^{2} - 2 y) - 1 \cdot - 3 = 0$

Factor $- 1$ out of $- (y^{2} - 2 y) - 1 (- 3)$ .

$- (y^{2} - 2 y - 3) = 0$

Factor.

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Factor $y^{2} - 2 y - 3$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Write the factored form using these integers.

$- ((y - 3) (y + 1)) = 0$

Remove unnecessary parentheses.

$- (y - 3) (y + 1) = 0$

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

$y - 3 = 0$

$y + 1 = 0$

Set $y - 3$ equal to $0$ and solve for $y$ .

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Set $y - 3$ equal to $0$ .

$y - 3 = 0$

Add $3$ to both sides of the equation.

$y = 3$

Set $y + 1$ equal to $0$ and solve for $y$ .

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Set $y + 1$ equal to $0$ .

$y + 1 = 0$

Subtract $1$ from both sides of the equation.

$y = - 1$

The final solution is all the values that make $- (y - 3) (y + 1) = 0$ true.

$y = 3, - 1$

Exclude the solutions that do not make $2 y - 3 = \sqrt{3 y^{2} - 10 y + 12}$ true.

$y = 3$