Given the equation
[tex]x^2=16x-65[/tex]
To find a and b, you have to find the roots, following the steps below.
Step 01: Write the equation in the general quadratic form.
The general quadratic form is ax²+bx+c=0.
Then, add -16x + 65 to both sides of the equation.
[tex]\begin{gathered} x^2-16+65=16x-65-16x+65 \\ x^2-16+65=0 \end{gathered}[/tex]
Step 2: Use the Bhaskara formula to find the roots.
The Bhaskara formula for a general equation is:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4\cdot a\cdot c}}{2\cdot a}[/tex]
In this exercise,
a = 1
b = -16
c = 65
Then,
[tex]\begin{gathered} x=\frac{-(-16)\pm\sqrt[]{(-16)^2-4\cdot1\cdot65}}{2\cdot1} \\ x=\frac{+16\pm\sqrt[]{256-260}}{2} \\ x=\frac{+16\pm\sqrt[]{-4}}{2} \end{gathered}[/tex]
√-4 can also be written as:
[tex]\sqrt[]{-4}=\sqrt[]{(4)\cdot(-1)}=\sqrt[]{4}\cdot\sqrt[]{-1}[/tex]
Knowing that i = √-1:
[tex]\sqrt[]{-4}=\sqrt[]{4}\cdot i[/tex]
Then:
[tex]\begin{gathered} x=\frac{+16\pm\sqrt[]{-4}}{2}=\frac{+16\pm\sqrt[]{4}\cdot i}{2} \\ x=\frac{16\pm2\cdot i}{2} \\ x=\frac{16}{2}\pm\frac{2}{2}\cdot i \\ x=8\pm i \end{gathered}[/tex]
The roots are:
8 + 1i
8 - 1i
So, the Answer is:
a = 8
b = 1
