Solve 5x+3=sqrt{3x+7}

$$5 x + 3 = \sqrt { 3 x + 7 }$$

Solve for x

$$x=-\frac{2}{25}=-0.08$$

Solution Steps

Square both sides of the equation.

$$\left(5x+3\right)^{2}=\left(\sqrt{3x+7}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5x+3\right)^{2}.

$$25x^{2}+30x+9=\left(\sqrt{3x+7}\right)^{2}$$

Calculate \sqrt{3x+7} to the power of 2 and get 3x+7.

$$25x^{2}+30x+9=3x+7$$

Subtract 3x from both sides.

$$25x^{2}+30x+9-3x=7$$

Combine 30x and -3x to get 27x.

$$25x^{2}+27x+9=7$$

Subtract 7 from both sides.

$$25x^{2}+27x+9-7=0$$

Subtract 7 from 9 to get 2.

$$25x^{2}+27x+2=0$$

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

$$a+b=27 ab=25\times 2=50$$

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 50.

$$1,50 2,25 5,10$$

Calculate the sum for each pair.

$$1+50=51 2+25=27 5+10=15$$

The solution is the pair that gives sum 27.

$$a=2 b=25$$

Rewrite 25x^{2}+27x+2 as \left(25x^{2}+2x\right)+\left(25x+2\right).

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

Factor out x in 25x^{2}+2x.

$$x\left(25x+2\right)+25x+2$$

Factor out common term 25x+2 by using distributive property.

$$\left(25x+2\right)\left(x+1\right)$$

To find equation solutions, solve 25x+2=0 and x+1=0.

$$x=-\frac{2}{25} x=-1$$

Substitute -\frac{2}{25} for x in the equation 5x+3=\sqrt{3x+7}.

$$5\left(-\frac{2}{25}\right)+3=\sqrt{3\left(-\frac{2}{25}\right)+7}$$

Simplify. The value x=-\frac{2}{25} satisfies the equation.

$$\frac{13}{5}=\frac{13}{5}$$

Substitute -1 for x in the equation 5x+3=\sqrt{3x+7}.

$$5\left(-1\right)+3=\sqrt{3\left(-1\right)+7}$$

Simplify. The value x=-1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-2=2$$

Equation 5x+3=\sqrt{3x+7} has a unique solution.

$$x=-\frac{2}{25}$$