Solve (7-y)^(2)+(y)^(2)=25 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$${ \left(7-y \right) }^{ 2 } + { y }^{ 2 } =25$$

Solve for y

$y=3$
$y=4$

Steps Using Factoring By Grouping

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using Factoring By Grouping

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(7-y\right)^{2}$.

$$49-14y+y^{2}+y^{2}=25$$

Combine $y^{2}$ and $y^{2}$ to get $2y^{2}$.

$$49-14y+2y^{2}=25$$

Subtract $25$ from both sides.

$$49-14y+2y^{2}-25=0$$

Subtract $25$ from $49$ to get $24$.

$$24-14y+2y^{2}=0$$

Divide both sides by $2$.

$$12-7y+y^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$y^{2}-7y+12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $y^{2}+ay+by+12$. To find $a$ and $b$, set up a system to be solved.

$$a+b=-7$$ $$ab=1\times 12=12$$

Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is negative, $a$ and $b$ are both negative. List all such integer pairs that give product $12$.

$$-1,-12$$ $$-2,-6$$ $$-3,-4$$

Calculate the sum for each pair.

$$-1-12=-13$$ $$-2-6=-8$$ $$-3-4=-7$$

The solution is the pair that gives sum $-7$.

$$a=-4$$ $$b=-3$$

Rewrite $y^{2}-7y+12$ as $\left(y^{2}-4y\right)+\left(-3y+12\right)$.

$$\left(y^{2}-4y\right)+\left(-3y+12\right)$$

Factor out $y$ in the first and $-3$ in the second group.

$$y\left(y-4\right)-3\left(y-4\right)$$

Factor out common term $y-4$ by using distributive property.

$$\left(y-4\right)\left(y-3\right)$$

To find equation solutions, solve $y-4=0$ and $y-3=0$.

$$y=4$$ $$y=3$$

Steps Using the Quadratic Formula

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(7-y\right)^{2}$.

$$49-14y+y^{2}+y^{2}=25$$

Combine $y^{2}$ and $y^{2}$ to get $2y^{2}$.

$$49-14y+2y^{2}=25$$

Subtract $25$ from both sides.

$$49-14y+2y^{2}-25=0$$

Subtract $25$ from $49$ to get $24$.

$$24-14y+2y^{2}=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.

$$2y^{2}-14y+24=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $2$ for $a$, $-14$ for $b$, and $24$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$y=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 2\times 24}}{2\times 2}$$

Square $-14$.

$$y=\frac{-\left(-14\right)±\sqrt{196-4\times 2\times 24}}{2\times 2}$$

Multiply $-4$ times $2$.

$$y=\frac{-\left(-14\right)±\sqrt{196-8\times 24}}{2\times 2}$$

Multiply $-8$ times $24$.

$$y=\frac{-\left(-14\right)±\sqrt{196-192}}{2\times 2}$$

Add $196$ to $-192$.

$$y=\frac{-\left(-14\right)±\sqrt{4}}{2\times 2}$$

Take the square root of $4$.

$$y=\frac{-\left(-14\right)±2}{2\times 2}$$

The opposite of $-14$ is $14$.

$$y=\frac{14±2}{2\times 2}$$

Multiply $2$ times $2$.

$$y=\frac{14±2}{4}$$

Now solve the equation $y=\frac{14±2}{4}$ when $±$ is plus. Add $14$ to $2$.

$$y=\frac{16}{4}$$

Divide $16$ by $4$.

$$y=4$$

Now solve the equation $y=\frac{14±2}{4}$ when $±$ is minus. Subtract $2$ from $14$.

$$y=\frac{12}{4}$$

Divide $12$ by $4$.

$$y=3$$

The equation is now solved.

$$y=4$$ $$y=3$$

Steps for Completing the Square

Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(7-y\right)^{2}$.

$$49-14y+y^{2}+y^{2}=25$$

Combine $y^{2}$ and $y^{2}$ to get $2y^{2}$.

$$49-14y+2y^{2}=25$$

Subtract $49$ from both sides.

$$-14y+2y^{2}=25-49$$

Subtract $49$ from $25$ to get $-24$.

$$-14y+2y^{2}=-24$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.

$$2y^{2}-14y=-24$$

Divide both sides by $2$.

$$\frac{2y^{2}-14y}{2}=-\frac{24}{2}$$

Dividing by $2$ undoes the multiplication by $2$.

$$y^{2}+\left(-\frac{14}{2}\right)y=-\frac{24}{2}$$

Divide $-14$ by $2$.

$$y^{2}-7y=-\frac{24}{2}$$

Divide $-24$ by $2$.

$$y^{2}-7y=-12$$

Divide $-7$, the coefficient of the $x$ term, by $2$ to get $-\frac{7}{2}$. Then add the square of $-\frac{7}{2}$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}-7y+\left(-\frac{7}{2}\right)^{2}=-12+\left(-\frac{7}{2}\right)^{2}$$

Square $-\frac{7}{2}$ by squaring both the numerator and the denominator of the fraction.

$$y^{2}-7y+\frac{49}{4}=-12+\frac{49}{4}$$

Add $-12$ to $\frac{49}{4}$.

$$y^{2}-7y+\frac{49}{4}=\frac{1}{4}$$

Factor $y^{2}-7y+\frac{49}{4}$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.

$$\left(y-\frac{7}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y-\frac{7}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$y-\frac{7}{2}=\frac{1}{2}$$ $$y-\frac{7}{2}=-\frac{1}{2}$$

Add $\frac{7}{2}$ to both sides of the equation.

$$y=4$$ $$y=3$$