Chapter 9.5, Problem 29E

To determine

To find the number of real solutions of the Quadratic equation.

Answer to Problem 29E

Two real solutions of the equation.

Explanation of Solution

Given:

Equation:

  $-\frac{1}{4}{x}^2+4x=-2$

Formula used:

Discriminant formula.

  $D=b^2-4ac$

Calculation:

Converting the equation in standard form

  $\frac{1}{4}{x}^2-4x+2=0$

On comparing the equation with $a{x}^2+bx+c$

  $a=\frac{1}{4},b=-4,c=2$

Finding the value of $D$

  $D=b^2-4ac$

Substituting the values,

  $D={(-4)}^2-4(\frac{1}{4})(2)$

Opening brackets,

  $D=16-2$

So,

  $D=14$

Since, $D$ is positive.

Therefore, there will two real solutions of the equation.