To find the solutions for the quadratic equation [tex]\( x^2 = 16x - 65 \)[/tex], we first need to rewrite it in standard form. The standard form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex].
Starting with the original equation:
[tex]\[ x^2 = 16x - 65 \][/tex]
We rearrange the terms to get:
[tex]\[ x^2 - 16x + 65 = 0 \][/tex]
Here, the coefficients are:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -16 \)[/tex]
- [tex]\( c = 65 \)[/tex]
To find the solutions, we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, we calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
[tex]\[ (-16)^2 - 4(1)(65) = 256 - 260 = -4 \][/tex]
Since the discriminant is negative, the equation has complex roots. Let's find the real and imaginary parts of these roots.
The real part of the roots is:
[tex]\[ \text{Real part} = \frac{-b}{2a} = \frac{16}{2} = 8 \][/tex]
The imaginary part of the roots is:
[tex]\[ \text{Imaginary part} = \frac{\sqrt{|\text{Discriminant}|}}{2a} = \frac{\sqrt{4}}{2} = 1 \][/tex]
So the solutions to the equation are in the form:
[tex]\[ x = a + bi \][/tex]
[tex]\[ x = a - bi \][/tex]
Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ x = 8 + 1i \][/tex]
[tex]\[ x = 8 - 1i \][/tex]
Therefore, the roots of the quadratic equation [tex]\( x^2 = 16x - 65 \)[/tex] are:
[tex]\[ x = 8 + 1i \][/tex]
[tex]\[ x = 8 - 1i \][/tex]
