Chapter 1.8, Problem 36E

Solution

After evaluating the discriminant of the quadratic equation $8z-10=z^2-7z+3$ , state the number and types of solution of it.

The discriminant of the quadratic equation is $D=173$ and it has two distinct real solutions $z=\left\{\frac{15+\sqrt{173}}{2},\frac{15-\sqrt{173}}{2}\right\}$ .

Given:

The quadratic equation is $8z-10=z^2-7z+3$ which can be written as $z^2-15z+13=0$ .

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 $b^2-4ac$ is called discriminant $\left(D\right)$ of the quadratic equation. Now to find the 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.

  $\left(1\right)$ If discriminant $D=0$ , then the quadratic equation a unique real solution.

Calculation:

Evaluate discriminant $\left(D\right)$ of the quadratic equation $z^2-15z+13=0$ .

  $\begin{array}{c}D={\left(-15\right)}^2-4\left(1\right)\left(13\right)\\ =173\end{array}$

Now discriminant $D=173$ is greater than zero, therefore the quadratic equation has two distinct real solution using clause $\left(1\right)$ of concept state above.

Now, the solutions for the quadratic equation $z^2-15z+13=0$ can be evaluated with the help of formula $\left(1\right)$ . Therefore,

  $\begin{array}{c}z=\frac{-b\pm \sqrt{D}}{2a}\mathrm{......}\left(\text{From }\left(1\right)\right)\\ =\frac{-\left(-15\right)\pm \sqrt{173}}{2\left(1\right)}\mathrm{......}\left(\begin{array}{l}\because z^2-15z+13=0\\ \text{ and }D=173\end{array}\right)\\ =\frac{15\pm \sqrt{173}}{2}\\ =\left\{\frac{15+\sqrt{173}}{2},\frac{15-\sqrt{173}}{2}\right\}\end{array}$

Conclusion:

The discriminant of the quadratic equation is $D=173$ and it has two distinct real solutions $z=\left\{\frac{15+\sqrt{173}}{2},\frac{15-\sqrt{173}}{2}\right\}$ .