Chapter 1.8, Problem 33E

Solution

After evaluating the discriminant of the quadratic equation $8p^2+8p+3=0$ , state the number and types of solution of it.

The discriminant of the quadratic equation is $D=-32$ and it has two imaginary solutions $p=\left\{\frac{-2+i\sqrt{2}}{4},\frac{-2-i\sqrt{2}}{4}\right\}$ .

Given:

The quadratic equation is $8p^2+8p+3=0$ .

Concept used:

Any quadratic equation of the form $a{x}^2+bx+c=0\text{ where }a\ne 0$ has the roots $x=\frac{-b\pm \sqrt{D}}{2a}\mathrm{......}\left(1\right)$

And the term $D=b^2-4ac$ is called discriminant $\left(D\right)$ of the quadratic equation. Now to find nature and number of roots of quadratic equation three cases arises.

  $\left(1\right)$ If discriminant $D>0$ , then the quadratic equation has two real and distinct solutions.

  $\left(2\right)$ If discriminant $D<0$ , then the quadratic equation has two imaginary distinct solutions which will be complex conjugate of each other.

  $\left(3\right)$ If discriminant $D=0$ , then the quadratic equation a unique real solution.

Calculation:

Evaluate discriminant $\left(D\right)$ of the quadratic equation $8p^2+8p+3=0$ .

  $\begin{array}{c}D={\left(8\right)}^2-4\left(8\right)\left(3\right)\\ =-32\end{array}$

Now discriminant $D=-32$ which is less than zero, therefore the quadratic equation has two imaginary solutions which will be complex conjugate to each other using clause $\left(2\right)$ of concept state above.

Now, the solution for the quadratic equation $8p^2+8p+3=0$ can be evaluated with the help of formula $\left(1\right)$ . Therefore,

  $\begin{array}{c}p=\frac{-b\pm \sqrt{D}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(8\right)\pm \sqrt{-32}}{2\left(8\right)}\mathrm{......}\left(\begin{array}{l}\because 8p^2+8p+3=0\\ \text{ and }D=-32\end{array}\right)\\ =\frac{-8\pm 4i\sqrt{2}}{16}\\ =\frac{-2\pm i\sqrt{2}}{4}\\ =\left\{\frac{-2+i\sqrt{2}}{4},\frac{-2-i\sqrt{2}}{4}\right\}\end{array}$

Conclusion:

The discriminant of the quadratic equation is $D=-32$ and it has two imaginary solutions $p=\left\{\frac{-2+i\sqrt{2}}{4},\frac{-2-i\sqrt{2}}{4}\right\}$ .