Solve y+1=sqrt{29-y}

$$y + 1 = \sqrt { 29 - y }$$

Solve for y

$$y=4$$

Solution Steps

Square both sides of the equation.

$$\left(y+1\right)^{2}=\left(\sqrt{29-y}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+1\right)^{2}.

$$y^{2}+2y+1=\left(\sqrt{29-y}\right)^{2}$$

Calculate \sqrt{29-y} to the power of 2 and get 29-y.

$$y^{2}+2y+1=29-y$$

Subtract 29 from both sides.

$$y^{2}+2y+1-29=-y$$

Subtract 29 from 1 to get -28.

$$y^{2}+2y-28=-y$$

Add y to both sides.

$$y^{2}+2y-28+y=0$$

Combine 2y and y to get 3y.

$$y^{2}+3y-28=0$$

To solve the equation, factor y^{2}+3y-28 using formula y^{2}+\left(a+b\right)y+ab=\left(y+a\right)\left(y+b\right). To find a and b, set up a system to be solved.

$$a+b=3 ab=-28$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -28.

$$-1,28 -2,14 -4,7$$

Calculate the sum for each pair.

$$-1+28=27 -2+14=12 -4+7=3$$

The solution is the pair that gives sum 3.

$$a=-4 b=7$$

Rewrite factored expression \left(y+a\right)\left(y+b\right) using the obtained values.

$$\left(y-4\right)\left(y+7\right)$$

To find equation solutions, solve y-4=0 and y+7=0.

$$y=4 y=-7$$

Substitute 4 for y in the equation y+1=\sqrt{29-y}.

$$4+1=\sqrt{29-4}$$

Simplify. The value y=4 satisfies the equation.

$$5=5$$

Substitute -7 for y in the equation y+1=\sqrt{29-y}.

$$-7+1=\sqrt{29-\left(-7\right)}$$

Simplify. The value y=-7 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-6=6$$

Equation y+1=\sqrt{29-y} has a unique solution.

$$y=4$$