If both p and q belong to the set {1, 2, 3, 4}, then how many equations of the form px 2+ qx + 1 = 0 will have real roots?
7
A quadratic equation is an equation of the form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are coefficients, and $a \neq 0$. The roots of a quadratic equation are the values of $x$ that satisfy the equation. These roots can be real or complex.
The nature of the roots of a quadratic equation is determined by its discriminant, which is given by the formula $\Delta = b^2 - 4ac$.
For a quadratic equation to have real roots, the discriminant must be greater than or equal to zero, i.e., $\Delta \ge 0$.
In the given equation $px^2 + qx + 1 = 0$, we have the coefficients:
For this equation to have real roots, the discriminant must satisfy $b^2 - 4ac \ge 0$. Substituting the coefficients, we get:
$\qquad q^2 - 4(p)(1) \ge 0$
$\qquad q^2 - 4p \ge 0$
$\qquad q^2 \ge 4p$
We are given that both $p$ and $q$ belong to the set $\{1, 2, 3, 4\}$. We need to find the number of pairs $(p, q)$ from this set that satisfy the condition $q^2 \ge 4p$.
We can list all possible pairs of $(p, q)$ where $p \in \{1, 2, 3, 4\}$ and $q \in \{1, 2, 3, 4\}$, and check if the inequality $q^2 \ge 4p$ holds for each pair.
| Value of p | Value of q | $q^2$ | $4p$ | Condition $q^2 \ge 4p$ | Real Roots? |
|---|---|---|---|---|---|
| 1 | 1 | $1^2 = 1$ | $4(1) = 4$ | $1 \ge 4$ (False) | No |
| 1 | 2 | $2^2 = 4$ | $4(1) = 4$ | $4 \ge 4$ (True) | Yes |
| 1 | 3 | $3^2 = 9$ | $4(1) = 4$ | $9 \ge 4$ (True) | Yes |
| 1 | 4 | $4^2 = 16$ | $4(1) = 4$ | $16 \ge 4$ (True) | Yes |
| 2 | 1 | $1^2 = 1$ | $4(2) = 8$ | $1 \ge 8$ (False) | No |
| 2 | 2 | $2^2 = 4$ | $4(2) = 8$ | $4 \ge 8$ (False) | No |
| 2 | 3 | $3^2 = 9$ | $4(2) = 8$ | $9 \ge 8$ (True) | Yes |
| 2 | 4 | $4^2 = 16$ | $4(2) = 8$ | $16 \ge 8$ (True) | Yes |
| 3 | 1 | $1^2 = 1$ | $4(3) = 12$ | $1 \ge 12$ (False) | No |
| 3 | 2 | $2^2 = 4$ | $4(3) = 12$ | $4 \ge 12$ (False) | No |
| 3 | 3 | $3^2 = 9$ | $4(3) = 12$ | $9 \ge 12$ (False) | No |
| 3 | 4 | $4^2 = 16$ | $4(3) = 12$ | $16 \ge 12$ (True) | Yes |
| 4 | 1 | $1^2 = 1$ | $4(4) = 16$ | $1 \ge 16$ (False) | No |
| 4 | 2 | $2^2 = 4$ | $4(4) = 16$ | $4 \ge 16$ (False) | No |
| 4 | 3 | $3^2 = 9$ | $4(4) = 16$ | $9 \ge 16$ (False) | No |
| 4 | 4 | $4^2 = 16$ | $4(4) = 16$ | $16 \ge 16$ (True) | Yes |
The pairs $(p, q)$ for which the equation $px^2 + qx + 1 = 0$ will have real roots are those that satisfy $q^2 \ge 4p$. From the table, these pairs are:
To find the total number of equations that will have real roots, we count the total number of valid pairs $(p, q)$ found above.
Total number of pairs = (Number of pairs for p=1) + (Number of pairs for p=2) + (Number of pairs for p=3) + (Number of pairs for p=4)
Total number of pairs = $3 + 2 + 1 + 1 = 7$
Therefore, there are 7 equations of the form $px^2 + qx + 1 = 0$ that will have real roots, given that $p, q \in \{1, 2, 3, 4\}$.
| Condition on Discriminant $\Delta = b^2 - 4ac$ | Nature of Roots |
|---|---|
| $\Delta > 0$ | Two distinct real roots |
| $\Delta = 0$ | One real root (repeated) |
| $\Delta < 0$ | Two complex roots |
| $\Delta \ge 0$ | Real roots (distinct or repeated) |
A quadratic equation is a polynomial equation of degree 2. The general form is $ax^2 + bx + c = 0$, where $a \neq 0$. The terms are $ax^2$ (quadratic term), $bx$ (linear term), and $c$ (constant term).
The roots of the quadratic equation can also be found using the quadratic formula:
$\qquad x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
The expression under the square root, $b^2 - 4ac$, is the discriminant ($\Delta$). As we saw, the value of the discriminant tells us whether the roots are real or complex.
In this problem, we were restricted to integer values for coefficients $p$ and $q$ from a small set. This allowed us to systematically list and check each possibility. For larger sets or continuous ranges of coefficients, graphical methods or further algebraic analysis might be needed to determine the conditions for real roots.
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