Chapter 7.3, Problem 4AYU

Find the real solutions of $x^2-x-1=0$.

To determine

To find: A Quadratic equation $x^2-x-1$ and get its real solutions.

Answer to Problem 4AYU

The solution set $=\left\{1+\surd \left(5\right)/2,1-\surd \left(5\right)/2\right\}$.

Explanation of Solution

Given:

A Quadratic equation $x^2–x–1=0$.

Formula used:

$x=-b\pm \surd \left(b^2–4ac\right)/2a$

Calculation:

There are various methods to solving a quadratic equation.

1. By factoring.
2. By using square root method.
3. By completing the square.
4. Using quadratic formula.

To find real solutions of Quadratic equation $x^2–x–1=0$ by factoring method.

Quadratic equation $x^2–x–1=0$ is in the standard form. The general equation of Quadratic equation $=a{x}^2+bx+c=0$.

Comparing $x^2–x–1=0$ and $a{x}^2+bx+c=0$ we get,

$a=1$, $b=-1$ and $c=-1$.

Checking for discriminant,

$b^2–4ac={\left(-1\right)}^2–4\times 1\times \left(-1\right)=1+4=5>0$

Therefore the solution will have 2 real solutions as degree of quadratic equation is 2 and discriminant is positive.

Using the formula,

$x=-b\pm \surd \left(b^2–4ac\right)/2a$

$x=-\left(-1\right)\pm \surd \left({\left(-1\right)}^2–4\times 1\times \left(-1\right)\right)/2\times 1$

$x=1\pm \surd \left(1+4\right)/2$

$x=1\pm \surd \left(5\right)/2$

$x=1+\surd \left(5\right)/2$ or $x=1-\surd \left(5\right)/2$

Therefore, the solution set $=\left\{1+\surd \left(5\right)/2,1-\surd \left(5\right)/2\right\}$.