Chapter 1.8, Problem 8GP

Solution

After evaluating the discriminant of the quadratic equation $4x^2+3x+12=3-3x$ , state the number and types of solution of it.

The discriminant of the quadratic equation is $D=-108$ and it has two imaginary solutions $x=\left\{\frac{-3+i3\sqrt{3}}{4},\frac{-3-i3\sqrt{3}}{4}\right\}$ .

Given:

The equation is $4x^2+3x+12=3-3x$ which can be written as $4x^2+6x+9=0$ .

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 $4x^2+6x+9=0$ .

  $\begin{array}{c}D={\left(6\right)}^2-4\left(4\right)\left(9\right)\\ =36-144\\ =-108\end{array}$

Now discriminant $D=-108$ 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 $4x^2+6x+9=0$ can be evaluated with the help of formula $\left(1\right)$ . Therefore,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{D}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(6\right)\pm \sqrt{-108}}{2\left(4\right)}\mathrm{......}\left(\begin{array}{l}\because 4x^2+6x+9=0\\ \text{ and }D=-108\end{array}\right)\\ =\frac{-6\pm i6\sqrt{3}}{8}\\ =\frac{-3\pm i3\sqrt{3}}{4}\\ =\left\{\frac{-3+i3\sqrt{3}}{4},\frac{-3-i3\sqrt{3}}{4}\right\}\end{array}$

Conclusion:

The discriminant of the quadratic equation is $D=-108$ and it has two imaginary solutions $x=\left\{\frac{-3+i3\sqrt{3}}{4},\frac{-3-i3\sqrt{3}}{4}\right\}$ .