Solve y+1=sqrt{11y-19}

$$y + 1 = \sqrt { 11 y - 19 }$$

Solve for y

$$y=5$$

$$y=4$$

Solution Steps

Square both sides of the equation.

$$\left(y+1\right)^{2}=\left(\sqrt{11y-19}\right)^{2}$$

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

$$y^{2}+2y+1=\left(\sqrt{11y-19}\right)^{2}$$

Calculate \sqrt{11y-19} to the power of 2 and get 11y-19.

$$y^{2}+2y+1=11y-19$$

Subtract 11y from both sides.

$$y^{2}+2y+1-11y=-19$$

Combine 2y and -11y to get -9y.

$$y^{2}-9y+1=-19$$

Add 19 to both sides.

$$y^{2}-9y+1+19=0$$

Add 1 and 19 to get 20.

$$y^{2}-9y+20=0$$

To solve the equation, factor y^{2}-9y+20 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.

$$a+b=-9 ab=20$$

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 20.

$$-1,-20 -2,-10 -4,-5$$

Calculate the sum for each pair.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

The solution is the pair that gives sum -9.

$$a=-5 b=-4$$

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

$$\left(y-5\right)\left(y-4\right)$$

To find equation solutions, solve y-5=0 and y-4=0.

$$y=5 y=4$$

Substitute 5 for y in the equation y+1=\sqrt{11y-19}.

$$5+1=\sqrt{11\times 5-19}$$

Simplify. The value y=5 satisfies the equation.

$$6=6$$

Substitute 4 for y in the equation y+1=\sqrt{11y-19}.

$$4+1=\sqrt{11\times 4-19}$$

Simplify. The value y=4 satisfies the equation.

$$5=5$$

List all solutions of y+1=\sqrt{11y-19}.

$$y=5 y=4$$