Question

Solve each equation. Express irrational solutions in exact form and as a decimal rounded to 3 decimal places. Verify your results using a graphing utility:$3^{x^2+x}=\sqrt{3}$

Short answer

The solution set of $x=\left\{\frac{-1-\sqrt{3}}{2},\frac{-1+\sqrt{3}}{2}\right\}\approx \left\{-1.366,0.366\right\}$

Step-by-step solution

Step 1. Given information 

An equation,$3^{x^2+x}=\sqrt{3}$

Step 2. Required to find 

Solve the given equation to find x.

Step 3. Solving the equation

Given, $3^{x^2+x}=\sqrt{3}$

role="math" localid="1646572989517" $\Rightarrow 3^{x^2+x}=3^{\frac{1}{2}}$

$\Rightarrow x^2+x=\frac{1}{2}$

$\Rightarrow x^2+x-0.5=0$

For quadratic equation, $a{x}^2+bx+c=0$, using quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-1\pm \sqrt{1^2-4\left(1\right)(-0.5)}}{2\left(1\right)}$

$x=\frac{-1\pm \sqrt{1+2}}{2}$

$x=\frac{-1\pm \sqrt{3}}{2}$

role="math" localid="1646573703954" $x=\frac{-1-\sqrt{3}}{2},x=\frac{-1+\sqrt{3}}{2}$

$x=-1.366,x=0.366$