Chapter 12, Problem 6CR

To determine

To define: The correct option of the expression $10y^2-y-3=0$ ,solve by quadratic formula and the options are,

  $\begin{array}{l}\text{a}.\left\{\frac{-3}{5},\frac{-1}{2}\right\}\\ \text{b}\text{.}\left\{\frac{3}{5},\frac{1}{2}\right\}\\ \text{c}.\left\{\frac{-3}{5},\frac{1}{2}\right\}\\ \text{d}\text{.}\text{}\left\{\frac{3}{5},\frac{-1}{2}\right\}\end{array}$

Answer to Problem 6CR

The correct option of expression $10y^2-y-3=0$ is $\text{d}\text{.}\text{}\left\{\frac{3}{5},\frac{-1}{2}\right\}$ .

Explanation of Solution

Given information:

The expression $10y^2-y-3=0$ and provided options are,

  $\begin{array}{l}\text{a}.\left\{\frac{-3}{5},\frac{-1}{2}\right\}\\ \text{b}\text{.}\left\{\frac{3}{5},\frac{1}{2}\right\}\\ \text{c}.\left\{\frac{-3}{5},\frac{1}{2}\right\}\\ \text{d}\text{.}\text{}\left\{\frac{3}{5},\frac{-1}{2}\right\}\end{array}$

The provided expression is $10y^2-y-3=0$ now solve for the value of $y$ by the quadratic formula.

Quadratic formula,

If $a{z}^2+bz+c=0$ , $a\ne 0$ and $b^2-4ac\ge 0$ .

Then $z=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Using the quadratic formula,

The values of $a=10,b=-1\text{and}c=-3$ put into the quadratic formula,

  $\begin{array}{l}y=\frac{-\left(-1\right)\pm \sqrt{1-4\cdot 10\cdot \left(-3\right)}}{2\cdot 10}\\ y=\frac{1\pm \sqrt{121}}{20}\\ y=\frac{1\pm 11}{20}\end{array}$

Furthermore,

  $\begin{array}{l}y=\frac{1\pm 11}{20}\\ y=\frac{3}{5}\text{or}y=\frac{-1}{2}\end{array}$

The value of $y$ is $\text{}\left\{\frac{3}{5},\frac{-1}{2}\right\}$ .

Thus, the option $\text{d}$ is correct.