Chapter 9, Problem 5PFA

To determine

To select the equation which has no real solution.

Answer to Problem 5PFA

Option B. $x^2-4x=-8$

Explanation of Solution

Given information :

Option A. $x^2-4x=-4$

Option B. $x^2-4x=-8$

Option C. $x^2+4x=-2$

Option D. $x^2+4x=2$

:

If the quadratic function $y=a{x}^2+bx+c$   has real solutions can be found out by using the quadratic formula:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The formula $b^2-4ac$ , also known as the discriminant can be used to determine the number of roots of solutions in a quadratic equation. There are three cases:

• $b^2-4ac<0$: The equation has 0 real solutions. The graph does not cross the  x -axis.
• $b^2-4ac=0$: The equation has 1 real solution. The graph touches the  x -axis at one point.
• $b^2-4ac>0$: The equation has 2 real solutions. The graph crosses through the  x -axis at two points.

Option A. $x^2-4x=-4$

Evaluating the discriminant.

  $b^2-4ac={(-4)}^2-4\times 1\times (4)$

Therefore, $b^2-4ac=0$ : The equation has 1 real solution.

Option B. $x^2-4x=-8$

Evaluating the discriminant.

  $b^2-4ac={(-4)}^2-4\times 1\times (8)$

Simplifying.

  $b^2-4ac=16-32$

Therefore, $b^2-4ac<0$ : The equation has 0 real solutions.

Option C. $x^2+4x=-2$

Evaluating the discriminant.

  $b^2-4ac={(4)}^2-4\times 1\times (2)$

Simplifying.

  $b^2-4ac=16-8$

Therefore, $b^2-4ac>0$ : The equation has 2 real solutions.

Option D. $x^2+4x=2$

Evaluating the discriminant.

  $b^2-4ac={(4)}^2-4\times 1\times (-2)$

Simplifying.

  $b^2-4ac=16+8$

Therefore, $b^2-4ac>0$ : The equation has 2 real solutions.