Chapter 5, Problem 12QC

To determine

To Find: The values of $x$ which satisfy the quadratic equation $2x^2+7x+3=0$ .

Answer to Problem 12QC

The values of $x$ are $-\frac{1}{2},-3$ .

Explanation of Solution

Given information:

The quadratic equation is $2x^2+7x+3=0$ .

Formula used:

The solution of a quadratic equation of the form $a{x}^2+bx+c=0$ ,where $a\ne 0$ is given by

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

Consider the quadratic equation $2x^2+7x+3=0$ .

Observe that the coefficient of $x^2,2\ne 0$ .

Recall that the solution of a quadratic equation of the form $a{x}^2+bx+c=0$ ,where $a\ne 0$ is given by

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Use it

  $\begin{array}{c}x=\frac{-7\pm \sqrt{7^2-4\times 2\times 3}}{2\times 2}\\ =\frac{-7\pm \sqrt{49-24}}{4}\\ =\frac{-7\pm \sqrt{25}}{4}\\ =\frac{-7\pm 5}{4}\end{array}$

Therefore $x=\frac{-7+5}{4}$ or $x=\frac{-7-5}{4}$ .That is $x=-\frac{1}{2}$ or $x=-3$

Thus, the values of $x$ are $-\frac{1}{2},-3$ .