Solve 4=sqrt{26-5x}+x

$$4 = \sqrt{ 26-5x } +x$$

Solve for x

$$x=-2$$

Solution Steps

Subtract x from both sides of the equation.

$$4-x=\sqrt{26-5x}$$

Square both sides of the equation.

$$\left(4-x\right)^{2}=\left(\sqrt{26-5x}\right)^{2}$$

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

$$16-8x+x^{2}=\left(\sqrt{26-5x}\right)^{2}$$

Calculate \sqrt{26-5x} to the power of 2 and get 26-5x.

$$16-8x+x^{2}=26-5x$$

Subtract 26 from both sides.

$$16-8x+x^{2}-26=-5x$$

Subtract 26 from 16 to get -10.

$$-10-8x+x^{2}=-5x$$

Add 5x to both sides.

$$-10-8x+x^{2}+5x=0$$

Combine -8x and 5x to get -3x.

$$-10-3x+x^{2}=0$$

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

$$x^{2}-3x-10=0$$

To solve the equation, factor x^{2}-3x-10 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.

$$a+b=-3 ab=-10$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -10.

$$1,-10 2,-5$$

Calculate the sum for each pair.

$$1-10=-9 2-5=-3$$

The solution is the pair that gives sum -3.

$$a=-5 b=2$$

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

$$\left(x-5\right)\left(x+2\right)$$

To find equation solutions, solve x-5=0 and x+2=0.

$$x=5 x=-2$$

Substitute 5 for x in the equation 4=\sqrt{26-5x}+x.

$$4=\sqrt{26-5\times 5}+5$$

Simplify. The value x=5 does not satisfy the equation.

$$4=6$$

Substitute -2 for x in the equation 4=\sqrt{26-5x}+x.

$$4=\sqrt{26-5\left(-2\right)}-2$$

Simplify. The value x=-2 satisfies the equation.

$$4=4$$

Equation 4-x=\sqrt{26-5x} has a unique solution.

$$x=-2$$