Chapter 1.9, Problem 10Q

Solution

After evaluating the discriminant of the quadratic equation $0=2x^2-3x+10$ , state the number and types of solution of it.

The discriminant of the quadratic equation is $-71$ and it has two imaginary solutions $x=\left\{\frac{3+i\sqrt{71}}{4},\frac{3-i\sqrt{71}}{4}\right\}$ .

Given:

The quadratic equation is $0=2x^2-3x+10$ which can be re-written as $2x^2-3x+10=0$ .

Concept used:

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

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 $2x^2-3x+10$ .

  $\begin{array}{c}D={\left(-3\right)}^2-4\left(2\right)\left(10\right)\mathrm{......}\left(2x^2-3x+10\right)\\ =9-80\\ =-71\end{array}$

Now discriminant $D=-71$ which is less than zero, therefore the quadratic equation has two imaginary solutions which will be complex conjugate to each other using clause $\left(2\right)$ in concept state above.

Now, the solution for the quadratic equation $2x^2-3x+10$ can be evaluated with the help of formula $\left(1\right)$ . Therefore,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{D}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(-3\right)\pm \sqrt{-71}}{2\left(2\right)}\mathrm{......}\left(\begin{array}{l}\because 2x^2-3x+10\\ \text{ and }D=-71\end{array}\right)\\ =\frac{3\pm i\sqrt{71}}{4}\\ =\left\{\frac{3+i\sqrt{71}}{4},\frac{3-i\sqrt{71}}{4}\right\}\end{array}$

Conclusion:

The discriminant of the quadratic equation is $-71$ and it has two imaginary solutions $x=\left\{\frac{3+i\sqrt{71}}{4},\frac{3-i\sqrt{71}}{4}\right\}$ .