Chapter 1.7, Problem 58E

Solution

To find: write a quadratic equation with real- number solutions.

The required quadratic equation is $x^2-8x-15=0$ .

Given:

Quadratic equation is to be considered.

Concept Used:

Method of solving quadratic equations.

Properties of square root: $a^2=b\Rightarrow a=\pm \sqrt{b}$

Algebraic identity: ${\left(a-b\right)}^2=a^2-2ab+b^2$

Calculation:

Suppose a quadratic equation: $x^2-8x-15=0$ with two real number solutions.

Now, Applying the factoring method:

  $x^2-8x-15=0$ .

Product of $-15\text{ and }1\text{ is }-15$

Now, factoring the product $-15$ :

  $\begin{array}{l}-1\text{ and 15 sum is (}-\text{1+15=14)}\\ \text{1 and }-\text{15 sum is (1}-\text{15=}-\text{14)}\\ -3\text{ and 5 sum is (}-3\text{+5=2)}\\ \text{3 and }-\text{5 sum is (3}-\text{5=}-2\text{)}\end{array}$

None of these sum equals to $-8$ . Hence factoring of this quadratic equation is not possible.

Here, applying the completing square method:

  $\begin{array}{c}{x}^2-8x-15=0\\ x^2-2\cdot 4\cdot x+4^2-4^2-15=0\\ {\left(x-4\right)}^2-16-15=0\left\{\because {\left(a-b\right)}^2=a^2-2ab+b^2\right\}\\ {\left(x-4\right)}^2=31\\ x-4=\pm \sqrt{31}\\ x=4\pm \sqrt{31}\end{array}$

Here, roots of the quadratic equation are $x=4+\sqrt{31},4-\sqrt{31}$ .

Therefore, the required quadratic equation is $x^2-8x-15=0$ .

Conclusion:

The required quadratic equation is $x^2-8x-15=0$ .