Solve dF(x)=(sqrt{x^2}+3x+5)

$$d F ( x ) = ( \sqrt { x ^ { 2 } } + 3 x + 5 )$$

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

Solution Steps

Subtract 3x+5 from both sides of the equation.

$$\frac{\mathrm{d}}{\mathrm{d}F}(x)-\left(3x+5\right)=\sqrt{x^{2}}$$

To find the opposite of 3x+5, find the opposite of each term.

$$\frac{\mathrm{d}}{\mathrm{d}F}(x)-3x-5=\sqrt{x^{2}}$$

Square both sides of the equation.

$$\left(\frac{\mathrm{d}}{\mathrm{d}F}(x)-3x-5\right)^{2}=\left(\sqrt{x^{2}}\right)^{2}$$

Square \frac{\mathrm{d}}{\mathrm{d}F}(x)-3x-5.

$$9x^{2}+30x+25=\left(\sqrt{x^{2}}\right)^{2}$$

Calculate \sqrt{x^{2}} to the power of 2 and get x^{2}.

$$9x^{2}+30x+25=x^{2}$$

Subtract x^{2} from both sides.

$$9x^{2}+30x+25-x^{2}=0$$

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$8x^{2}+30x+25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx+25. To find a and b, set up a system to be solved.

$$a+b=30 ab=8\times 25=200$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 200.

$$1,200 2,100 4,50 5,40 8,25 10,20$$

Calculate the sum for each pair.

$$1+200=201 2+100=102 4+50=54 5+40=45 8+25=33 10+20=30$$

The solution is the pair that gives sum 30.

$$a=10 b=20$$

Rewrite 8x^{2}+30x+25 as \left(8x^{2}+10x\right)+\left(20x+25\right).

$$\left(8x^{2}+10x\right)+\left(20x+25\right)$$

Factor out 2x in the first and 5 in the second group.

$$2x\left(4x+5\right)+5\left(4x+5\right)$$

Factor out common term 4x+5 by using distributive property.

$$\left(4x+5\right)\left(2x+5\right)$$

To find equation solutions, solve 4x+5=0 and 2x+5=0.

$$x=-\frac{5}{4} x=-\frac{5}{2}$$

Substitute -\frac{5}{4} for x in the equation \frac{\mathrm{d}}{\mathrm{d}F}(x)=\sqrt{x^{2}}+3x+5.

$$\frac{\mathrm{d}}{\mathrm{d}F}(-\frac{5}{4})=\sqrt{\left(-\frac{5}{4}\right)^{2}}+3\left(-\frac{5}{4}\right)+5$$

Simplify. The value x=-\frac{5}{4} does not satisfy the equation.

$$0=\frac{5}{2}$$

Substitute -\frac{5}{2} for x in the equation \frac{\mathrm{d}}{\mathrm{d}F}(x)=\sqrt{x^{2}}+3x+5.

$$\frac{\mathrm{d}}{\mathrm{d}F}(-\frac{5}{2})=\sqrt{\left(-\frac{5}{2}\right)^{2}}+3\left(-\frac{5}{2}\right)+5$$

Simplify. The value x=-\frac{5}{2} satisfies the equation.

$$0=0$$

Equation \frac{\mathrm{d}}{\mathrm{d}F}(x)-3x-5=\sqrt{x^{2}} has a unique solution.

$$x=-\frac{5}{2}$$

Solve for F

$$F\in \mathrm{R}$$

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$