Chapter 4.7, Problem 9LC

Solution

To Explain: Whether if two discriminants are equal, the solutions to the two equations are same. Explain the error and give an example of two quadratic equations.

No, if two discriminants are equal, the solutions to the two equations are not same.

Given:

If whether two discriminants are equal, the solutions to the two equations are same. Explain the error and give an example of two quadratic equations.

Explanation:

When there is one solution to an equation, the discriminants are all zero and thus equal, but if there are two or more solutions, they may or may not be equal.

For example,

  $x^2+6x+9=0$

Compared it with $a{x}^2+bx+c=0$

Thus, the values are $a=1,b=6,$ and $c=9$

The discriminate of the equation (1) is:

  $\begin{array}{l}=b^2-4ac\\ =6^2-4\left(1\right)\left(9\right)\\ =36-36\\ =0\end{array}$

The roots of quadratic equation $a{x}^2+bx+c=0$ is:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $a=1,b=6,$ and $c=9$ :

  $\begin{array}{c}x=\frac{-6\pm \sqrt{0}}{2\left(1\right)}\\ =\frac{-6}{2}\\ =-3\end{array}$

There is only one real solution for the discriminate with zero.

Consider another example: $x^2-4x+4=0$

Compared the above equation with $a{x}^2+bx+c=0$

Thus, the values are $a=1,b=-4,$ and $c=4$

The discriminate of the equation is:

  $\begin{array}{l}=b^2-4ac\\ ={\left(-4\right)}^2-4\left(1\right)\left(4\right)\\ =16-16\\ =0\end{array}$

The roots of quadratic equation $a{x}^2+bx+c=0$

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $a=1,b=-4,$ and $c=4$ :

  $\begin{array}{l}x=\frac{-\left(-4\right)\pm \sqrt{0}}{2\left(1\right)}\\ x=\frac{4}{2}\\ x=2\end{array}$

There is only one real solution for the discriminate with zero.

There is a discriminate zero in equations $\left(1\right)$ and $\left(3\right)$ , but the solutions of equations $\left(1\right)$ and $\left(3\right)$ are different.