Algebra Examples

$9 y^{2} - 30 y + 25 = 0$

Step 1

Solve for $y$ .

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Step 1.1

Factor using the perfect square rule.

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Step 1.1.1

Rewrite $9 y^{2}$ as ${(3 y)}^{2}$ .

${(3 y)}^{2} - 30 y + 25 = 0$

Step 1.1.2

Rewrite $25$ as $5^{2}$ .

${(3 y)}^{2} - 30 y + 5^{2} = 0$

Step 1.1.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$30 y = 2 \cdot (3 y) \cdot 5$

Step 1.1.4

Rewrite the polynomial.

${(3 y)}^{2} - 2 \cdot (3 y) \cdot 5 + 5^{2} = 0$

Step 1.1.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 3 y$ and $b = 5$ .

${(3 y - 5)}^{2} = 0$

Step 1.2

Set the $3 y - 5$ equal to $0$ .

$3 y - 5 = 0$

Step 1.3

Solve for $y$ .

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Step 1.3.1

Add $5$ to both sides of the equation.

$3 y = 5$

Step 1.3.2

Divide each term in $3 y = 5$ by $3$ and simplify.

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Step 1.3.2.1

Divide each term in $3 y = 5$ by $3$ .

$\frac{3 y}{3} = \frac{5}{3}$

Step 1.3.2.2

Simplify the left side.

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Step 1.3.2.2.1

Cancel the common factor of $3$ .

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Step 1.3.2.2.1.1

Cancel the common factor.

$\frac{3 y}{3} = \frac{5}{3}$

Step 1.3.2.2.1.2

Divide $y$ by $1$ .

$y = \frac{5}{3}$

Step 2

Set $0$ equal to $0$ .

$\frac{5}{3} = 0$

Step 3

Solve for $x$ .

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Step 3.1

Set the numerator equal to zero.

$5 = 0$

Step 3.2

Since $5 \neq 0$ , there are no solutions.

No solution