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What are the solutions of this quadratic equation?
[tex]\[ x^2 = 16x - 65 \][/tex]

Substitute the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] to complete the solutions:
[tex]\[
\begin{array}{l}
x = a + b i \\
x = a - b i
\end{array}
\][/tex]


Sagot :

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]
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