Chapter A.6, Problem 132AYU

To determine

Make up three quadratic equations.

Answer to Problem 132AYU

Three quadratic equations has been explained.

Explanation of Solution

Given information:

Make up three quadratic equations: one having two distinct solutions, and one having exactly one real solution.

Calculation:

Create three different quadratic equations: one that has two unequal solutions, one with no real solution, and one with exactly one real solution.

To create a quadratic equation of the form $a{x}^2+bx+c=0$ , we could consider the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

If the value of the discriminant $b^2-4ac$ is positive, there are two real unequal solutions. If the value of the discriminant is $0$ , there is one repeated solution. If the value is negative, there is no real solution.

Consider the equation $x^2+x-3=0$ . Here $a=1$ , $b=1$ and $c=-3$ . Plug these values into the descriminant to determine the number and type of solutions

  $b^2-4ac=1^2-4(1)(-3)$

  $=1+12$

  $=13$

Because the discriminant is positive, the equation $x^2+x-3=0$ has two unequal real solutions.

Consider the equation $x^2+x+3=0$ . Here $a=1$ , $b=1$ and $c=3$ . Plug these values into the discriminant to determine the number and type of solutions

  $b^2-4ac=1^2-4(1)(3)$

  $=1-12$

  $=-11$

Because the discriminant is negative, the equation $x^2+x+3=0$ has no real solution.

Consider the equation $x^2+8x+16=0$ . Here $a=1$ , $b=8$ and $c=16$ . Plug these values into the discriminant to determine the number and type of solutions.

  $b^2-4ac=8^2-4(1)(16)$

  $=64-64$

  $=0$

Because the discriminant is $0$ , the equation $x^2+8x+16=0$ has one repeated real solution.

Hence three quadratic equations has been explained.