Solve x-4=sqrt{6x-17}

$$x - 4 = \sqrt { 6 x - 17 }$$

Solve for x

$$x=11$$

Solution Steps

Square both sides of the equation.

$$\left(x-4\right)^{2}=\left(\sqrt{6x-17}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16=\left(\sqrt{6x-17}\right)^{2}$$

Calculate \sqrt{6x-17} to the power of 2 and get 6x-17.

$$x^{2}-8x+16=6x-17$$

Subtract 6x from both sides.

$$x^{2}-8x+16-6x=-17$$

Combine -8x and -6x to get -14x.

$$x^{2}-14x+16=-17$$

Add 17 to both sides.

$$x^{2}-14x+16+17=0$$

Add 16 and 17 to get 33.

$$x^{2}-14x+33=0$$

To solve the equation, factor x^{2}-14x+33 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-14 ab=33$$

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 33.

$$-1,-33 -3,-11$$

Calculate the sum for each pair.

$$-1-33=-34 -3-11=-14$$

The solution is the pair that gives sum -14.

$$a=-11 b=-3$$

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

$$\left(x-11\right)\left(x-3\right)$$

To find equation solutions, solve x-11=0 and x-3=0.

$$x=11 x=3$$

Substitute 11 for x in the equation x-4=\sqrt{6x-17}.

$$11-4=\sqrt{6\times 11-17}$$

Simplify. The value x=11 satisfies the equation.

$$7=7$$

Substitute 3 for x in the equation x-4=\sqrt{6x-17}.

$$3-4=\sqrt{6\times 3-17}$$

Simplify. The value x=3 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-1=1$$

Equation x-4=\sqrt{6x-17} has a unique solution.

$$x=11$$