Math Problem Statement

Determine if the following equations have two real solutions, one real solution, or two imaginary solutions.

Solution

Let's analyze the given quadratic equations to determine whether they have two real solutions, one real solution, or two imaginary solutions. This can be done by calculating the discriminant Δ\Delta for each equation.

For a quadratic equation of the form:

ax2+bx+c=0ax^2 + bx + c = 0

The discriminant Δ\Delta is given by:

Δ=b24ac\Delta = b^2 - 4ac

  • If Δ>0\Delta > 0, there are two distinct real solutions.
  • If Δ=0\Delta = 0, there is one real solution.
  • If Δ<0\Delta < 0, there are two imaginary (complex) solutions.

Let’s solve each equation step by step.


11. 3x2240=03x^2 - 240 = 0

Rewrite as: 3x2=2403x^2 = 240 This is a standard quadratic equation, and we can directly solve for the discriminant:

  • a=3a = 3
  • b=0b = 0
  • c=240c = -240

Discriminant Δ=024(3)(240)=2880\Delta = 0^2 - 4(3)(-240) = 2880

Since Δ>0\Delta > 0, two real solutions.


12. 5x2=60-5x^2 = 60

Rewrite as: 5x260=0-5x^2 - 60 = 0

  • a=5a = -5
  • b=0b = 0
  • c=60c = -60

Discriminant Δ=024(5)(60)=1200\Delta = 0^2 - 4(-5)(-60) = -1200

Since Δ<0\Delta < 0, two imaginary solutions.


13. x24+9=0\frac{x^2}{4} + 9 = 0

Rewrite as: x24=9\frac{x^2}{4} = -9 or x2=36x^2 = -36

  • This gives imaginary solutions directly because x2=36x^2 = -36.

Thus, two imaginary solutions.


14. 7(x+1)224=07(x + 1)^2 - 24 = 0

Rewrite as: 7(x+1)2=247(x + 1)^2 = 24, or (x+1)2=247(x + 1)^2 = \frac{24}{7}

  • Taking the square root, we find two real roots.

Thus, two real solutions.


15. 2(3x2+16)=322(3x^2 + 16) = 32

Simplify to: 3x2+16=163x^2 + 16 = 16, then 3x2=03x^2 = 0

  • This simplifies to x=0x = 0.

Thus, one real solution.


16. x22+12=0-\frac{x^2}{2} + 12 = 0

Rewrite as: x22=12-\frac{x^2}{2} = -12, or x2=24x^2 = 24

  • This gives two real roots since x=±24x = \pm \sqrt{24}.

Thus, two real solutions.


Summary:

  • 11: Two real solutions
  • 12: Two imaginary solutions
  • 13: Two imaginary solutions
  • 14: Two real solutions
  • 15: One real solution
  • 16: Two real solutions

Would you like more details or have any questions about these calculations?

Here are five related questions:

  1. What is the significance of the discriminant in quadratic equations?
  2. How can you factor a quadratic equation to find its roots?
  3. Can a quadratic equation have both real and imaginary solutions?
  4. How do you solve a quadratic equation by completing the square?
  5. What is the quadratic formula and when is it used?

Tip: When solving quadratic equations, always check if they can be factored easily before using the quadratic formula!

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Discriminant

Formulas

Discriminant formula: Δ = b^2 - 4ac
Quadratic equation standard form: ax^2 + bx + c = 0

Theorems

Quadratic Discriminant Theorem

Suitable Grade Level

Grades 9-12