Answer:
238
Step-by-step explanation:
Given equation: [tex]a^2+1=6a[/tex]
Rearrange so that it is equal to zero:
[tex]\implies a^2-6a+1=0[/tex]
Quadratic formula
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when}\:ax^2+bx+c=0[/tex]
Using the quadratic formula to solve the given equation:
[tex]\implies a=\dfrac{-(-6) \pm \sqrt{(-6)^2-4(1)(1)} }{2(1)}[/tex]
[tex]\implies a=\dfrac{6 \pm \sqrt{32}}{2}[/tex]
[tex]\implies a=3 \pm 2\sqrt{2}[/tex]
[tex]\implies a^2=17 \pm 12\sqrt{2}[/tex]
Substituting this into [tex]7\left(a^2+\dfrac{1}{a^2}\right)[/tex]
[tex]\implies 7\left(17 \pm 12\sqrt{2}+\dfrac{1}{17 \pm 12\sqrt{2}}\right)[/tex]
For [tex]17+12\sqrt{2}[/tex]:
[tex]\implies 7\left(17+12\sqrt{2}+\dfrac{1}{17+12\sqrt{2}}\right)[/tex]
[tex]\implies 7\left(\dfrac{(17+12\sqrt{2})^2}{17+12\sqrt{2}}+\dfrac{1}{17+12\sqrt{2}}\right)[/tex]
[tex]\implies 7\left(\dfrac{(17+12\sqrt{2})^2+1}{17+12\sqrt{2}}\right)[/tex]
[tex]\implies 7\left(\dfrac{289+408\sqrt{2}+288+1}{17+12\sqrt{2}}\right)[/tex]
[tex]\implies 7\left(\dfrac{578+408\sqrt{2}}{17+12\sqrt{2}}\right)[/tex]
[tex]\implies 7\left(\dfrac{34(17+12\sqrt{2})}{17+12\sqrt{2}}\right)[/tex]
[tex]\implies 7(34)[/tex]
[tex]\implies 238[/tex]
For [tex]17-12\sqrt{2}[/tex]:
[tex]\implies 7\left(17-12\sqrt{2}+\dfrac{1}{17-12\sqrt{2}}\right)[/tex]
[tex]\implies 7\left(\dfrac{(17-12\sqrt{2})^2}{17-12\sqrt{2}}+\dfrac{1}{17-12\sqrt{2}}\right)[/tex]
[tex]\implies 7\left(\dfrac{(17-12\sqrt{2})^2+1}{17-12\sqrt{2}}\right)[/tex]
[tex]\implies 7\left(\dfrac{289-408\sqrt{2}+288+1}{17-12\sqrt{2}}\right)[/tex]
[tex]\implies 7\left(\dfrac{578-408\sqrt{2}}{17-12\sqrt{2}}\right)[/tex]
[tex]\implies 7\left(\dfrac{34(17-12\sqrt{2})}{17-12\sqrt{2}}\right)[/tex]
[tex]\implies 7(34)[/tex]
[tex]\implies 238[/tex]
