Math Problem Statement

If both p and q belong the set (1, 2, 3, 4, 5, 6, 7, 8), then how many equations of the form px² + qx + 2 = 0 have real roots?

Solution

To determine how many equations of the form px2+qx+2=0px^2 + qx + 2 = 0 have real roots where pp and qq belong to the set {1,2,3,4,5,6,7,8}\{1, 2, 3, 4, 5, 6, 7, 8\}, we need to ensure that the discriminant of the quadratic equation is non-negative. Recall that for a quadratic equation of the form ax2+bx+c=0ax^2 + bx + c = 0, the discriminant is given by:

D=b24acD = b^2 - 4ac

In our case, we have a=pa = p, b=qb = q, and c=2c = 2. Therefore, the discriminant becomes:

D=q24×p×2=q28pD = q^2 - 4 \times p \times 2 = q^2 - 8p

For the equation to have real roots, the discriminant must be greater than or equal to zero:

q28p0q^2 - 8p \geq 0

Rearranging this inequality gives:

q28pq^2 \geq 8p

Now, we will check all combinations of pp and qq from the set {1,2,3,4,5,6,7,8}\{1, 2, 3, 4, 5, 6, 7, 8\} to see when this condition holds:

  1. If p=1p = 1, then q28q^2 \geq 8. The possible values of qq are 3,4,5,6,7,83, 4, 5, 6, 7, 8 (since q2q^2 must be at least 8).
  2. If p=2p = 2, then q216q^2 \geq 16. The possible values of qq are 4,5,6,7,84, 5, 6, 7, 8.
  3. If p=3p = 3, then q224q^2 \geq 24. The possible values of qq are 5,6,7,85, 6, 7, 8.
  4. If p=4p = 4, then q232q^2 \geq 32. The possible values of qq are 6,7,86, 7, 8.
  5. If p=5p = 5, then q240q^2 \geq 40. The possible values of qq are 7,87, 8.
  6. If p=6p = 6, then q248q^2 \geq 48. The possible values of qq are 7,87, 8.
  7. If p=7p = 7, then q256q^2 \geq 56. The possible values of qq are 88.
  8. If p=8p = 8, then q264q^2 \geq 64. The possible values of qq are 88.

Next, we count all valid combinations:

  • p=1p = 1: 6 values of qq (3, 4, 5, 6, 7, 8)
  • p=2p = 2: 5 values of qq (4, 5, 6, 7, 8)
  • p=3p = 3: 4 values of qq (5, 6, 7, 8)
  • p=4p = 4: 3 values of qq (6, 7, 8)
  • p=5p = 5: 2 values of qq (7, 8)
  • p=6p = 6: 2 values of qq (7, 8)
  • p=7p = 7: 1 value of qq (8)
  • p=8p = 8: 1 value of qq (8)

Total number of combinations is:

6+5+4+3+2+2+1+1=246 + 5 + 4 + 3 + 2 + 2 + 1 + 1 = 24

Hence, there are 24 equations of the form px2+qx+2=0px^2 + qx + 2 = 0 that have real roots.

Would you like to see further details or have any questions? Here are some related problems you might find interesting:

  1. What if the constant term is changed to 1? How many such equations have real roots?
  2. Explore the discriminant condition for different ranges of pp and qq.
  3. What happens if the values of pp and qq are restricted to odd numbers?
  4. Consider the case when the coefficient of x2x^2 is allowed to be negative.
  5. How many equations have exactly one real root?

Tip: For a quadratic equation to have real roots, always check if its discriminant is non-negative.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Discriminant of a Quadratic Equation

Formulas

General form of quadratic equation: ax² + bx + c = 0
Discriminant formula: D = b² - 4ac
Condition for real roots: D ≥ 0

Theorems

Discriminant theorem for real roots of a quadratic equation

Suitable Grade Level

Grades 10-12