Chapter 4, Problem 19AR

To determine

To solve the given equation, $3x^2+3x=4$ by factoring or quadratic formula.

Answer to Problem 19AR

  $x=\frac{-3+\sqrt{57}}{6},\frac{-3-\sqrt{57}}{6}$

Explanation of Solution

Given:

The equation is, $3x^2+3x=4$

Calculation:

We can solve the given equation by using quadratic formula.

That is,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Now, write the equation as,

  $\begin{array}{l}3x^2+3x=4\\ 3x^2+3x-4=0\end{array}$

Which is in the form of $a{x}^2+bx+c=0$ . Where, $a=3,b=3\text{and c = -4}$ .

Now, Substitute the values in the formula then we get that,

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-3\pm \sqrt{3^2-4\left(3\right)\left(-4\right)}}{2\left(3\right)}\\ x=\frac{-3\pm \sqrt{9+48}}{6}\\ x=\frac{-3\pm \sqrt{57}}{6}\\ x=\frac{-3+\sqrt{57}}{6},\frac{-3-\sqrt{57}}{6}\end{array}$

Therefore, the roots for the given equation, $3x^2+3x=4$ are $x=\frac{-3+\sqrt{57}}{6},\frac{-3-\sqrt{57}}{6}$ .