Chapter 9.6, Problem 55HP

To determine

Wheather there are two, one, or no real solution

Answer to Problem 55HP

  $x=1+\frac{\sqrt{30}}{6},\text{ and x}=1-\frac{\sqrt{30}}{6}$

Explanation of Solution

Given:

The graph of a quadratic function intersects the x-axis twice

Concept Used:

Quadratic formula: The roots of a quadratic equation $a{x}^2+bx+c=0;\text{ where }a\ne 0$ are:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

Let’s assume a equation $6x^2-12x+1=0$

In order to solve the given equation $6x^2-12x+1=0$ , since it is in the descending order of exponents, so comparing with the general equation $a{x}^2+bx+c=0$ , the value of the coefficient a, b, and c is given by

  $a=6,b=-12,\text{ and }c=1$

Now, above quadratic equation $6y^2-12y+1=0$ can be solved using quadratic formula as shown below,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(-12\right)\pm \sqrt{{\left(-12\right)}^2-4\left(6\right)\left(1\right)}}{2\left(6\right)}\\ =\frac{12\pm \sqrt{144-24}}{12}\\ =\frac{12\pm \sqrt{120}}{12}\\ =\frac{12\pm \sqrt{120}}{12}\\ =\frac{12+\sqrt{4\times 30}}{12}\\ =\frac{12+2\sqrt{30}}{12}\\ \Rightarrow x=\frac{12}{12}+\frac{2\sqrt{30}}{12}\\ \Rightarrow x=1+\frac{\sqrt{30}}{6},\text{ and x}=1-\frac{\sqrt{30}}{6}\\ \end{array}$

Thus, the solutions of the given question are, $x=1+\frac{\sqrt{30}}{6},\text{ and x}=1-\frac{\sqrt{30}}{6}$ .

Graph of the equation is given below