Question

1. solve for j. 3j² - 46j + 15 = 0 1) write each solution as an integer, proper - fraction, or improper fraction in simplest form. if there are multiple solutions, separate them with commas. j =

Explanation:

Step1: Identify coefficients

For the quadratic equation $3j^{2}-46j + 15=0$, we have $a = 3$, $b=-46$, $c = 15$.

Step2: Use quadratic formula

The quadratic formula is $j=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute the values:
\[

$$\begin{align*} j&=\frac{-(-46)\pm\sqrt{(-46)^{2}-4\times3\times15}}{2\times3}\\ &=\frac{46\pm\sqrt{2116 - 180}}{6}\\ &=\frac{46\pm\sqrt{1936}}{6}\\ &=\frac{46\pm44}{6} \end{align*}$$

\]

Step3: Calculate two solutions

For the plus - sign: $j=\frac{46 + 44}{6}=\frac{90}{6}=15$.
For the minus - sign: $j=\frac{46-44}{6}=\frac{2}{6}=\frac{1}{3}$.

Answer:

$15,\frac{1}{3}$