Given the equation:
[tex]x^2-10x+26=8[/tex]
Let's find the solution using the quadratic formula.
Equate to zero by subtracting x from both sides of the equation:
[tex]\begin{gathered} x^2-10x+26-8=8-8 \\ \\ x^2-10x+18=0 \end{gathered}[/tex]
Apply the quadratic formula:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Where:
a = 1
b = -10
c = 18
Plug in the values in the equation and solve for x.
We have:
[tex]\begin{gathered} x=\frac{--10\pm\sqrt{-10^2-4(1)(18)}}{2(1)} \\ \\ x=\frac{10\pm\sqrt{100-72}}{2} \\ \\ x=\frac{10\pm\sqrt{28}}{2} \end{gathered}[/tex]
Solving further:
[tex]\begin{gathered} x=\frac{10+\sqrt{7*4}}{2} \\ \\ x=\frac{10+\sqrt{7*2^2}}{2} \\ \\ x=\frac{10+2\sqrt{7}}{2} \\ \\ x=\frac{10}{2}\pm\frac{2\sqrt{7}}{2} \\ \\ x=5\pm\sqrt{7} \\ \\ x=5-\sqrt{7} \\ x=5+\sqrt{7} \end{gathered}[/tex]
Therefore, the values of a and b are:
a = 5
b = 7
• ANSWER:
a = 5
b = 7
