what is the value of "a" if the equation (9+i)x^2+(a-i)x+36-2i=0 has
real roots?
To determine the value of "a" such that the equation has real roots, we need to find the discriminant of the quadratic equation.
The discriminant (D) is a mathematical expression used to determine the nature of the roots of a quadratic equation. For a quadratic equation of the form ax^2 + bx + c = 0, the discriminant is calculated as:
D = b^2 - 4ac
In our given equation (9+i)x^2 + (a-i)x + 36-2i = 0, we can identify the coefficients as:
a = (9+i)
b = (a-i)
c = (36-2i)
Substituting these values into the discriminant formula, we have:
D = [(a - i)^2] - 4(9+i)(36-2i)
Expanding and simplifying, we get:
D = (a^2 - 2ai + i^2) - 4(324 - 2ai + 72i - 4i^2)
= a^2 - 2ai - 1 + 1296a - 8ai - 144i + 16
Combining like terms, we have:
D = a^2 - 1 + 1296a - 144i - 10ai + 16
Since we want the equation to have real roots, the discriminant (D) must be non-negative (D ≥ 0). To ensure this, the imaginary part (-144i - 10ai) should be zero.
Hence, setting the imaginary part equal to zero:
-144i - 10ai = 0
Simplifying, we get:
-144i = 10ai
divide both sides by i
-144 = 10a
solve for a
a = -144/10
a = -14.4
So, the value of "a" for the equation to have real roots is -14.4.