Chapter 12, Problem 8CT

To determine

To calculate: The number of real roots of equation $x^2+4x-4=0$ .

Answer to Problem 8CT

The number of real roots of equation $x^2+4x-4=0$ is 2.

Explanation of Solution

Given information:

The expression is given as:

  $x^2+4x-4=0$

Formula used:

If $a{z}^2+bz+c=0$ , $a\ne 0$

Then number of real roots are,

If $b^2-4ac>0$ then $a{z}^2+bz+c=0$ have 2 different real roots.

If $b^2-4ac=0$ then $a{z}^2+bz+c=0$ have 2 repeated real roots.

If $b^2-4ac<0$ then $a{z}^2+bz+c=0$ have no real roots.

Calculation:

Using the rule of real roots on discriminant of quadratic equation,

The values of $a=1,b=4\text{and}c=-4$ put into the discriminant of quadratic equation,

  $\begin{array}{l}{b}^2-4ac\\ ={\left(4\right)}^2-4\cdot 1\cdot \left(-4\right)\\ =16+16\\ =32\end{array}$

The discriminant of quadratic equation is negative.

So, by this rule If $b^2-4ac>0$ then $a{z}^2+bz+c=0$ have 2 different real roots, there are 2 real roots of the equation.