To determine the value of [tex]\( p \)[/tex] that ensures the quadratic equation [tex]\( x^2 + px + 49 = 0 \)[/tex] has exactly one real solution, we need to use the concept of the discriminant.
A quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex] has its solutions determined by the discriminant [tex]\(\Delta\)[/tex], which is given by the formula:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
For our equation [tex]\( x^2 + px + 49 = 0 \)[/tex], we have:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = p \)[/tex]
- [tex]\( c = 49 \)[/tex]
The equation will have exactly one real solution if and only if the discriminant [tex]\(\Delta\)[/tex] equals zero. Therefore, we set up the equation:
[tex]\[ \Delta = p^2 - 4 \cdot 1 \cdot 49 = 0 \][/tex]
Simplifying this, we have:
[tex]\[ p^2 - 196 = 0 \][/tex]
Next, we solve for [tex]\( p \)[/tex]:
[tex]\[ p^2 = 196 \][/tex]
Taking the square root of both sides, we get:
[tex]\[ p = \sqrt{196} \][/tex]
Since [tex]\( p \)[/tex] must be greater than zero, we take the positive square root:
[tex]\[ p = 14 \][/tex]
To verify, let's check the discriminant:
[tex]\[ \Delta = 14^2 - 4 \cdot 1 \cdot 49 \][/tex]
[tex]\[ \Delta = 196 - 196 \][/tex]
[tex]\[ \Delta = 0 \][/tex]
Thus, the value of [tex]\( p \)[/tex] that ensures the quadratic equation [tex]\( x^2 + px + 49 = 0 \)[/tex] has one real solution is:
[tex]\[ \boxed{14} \][/tex]
