+0

# Algebra

0
179
1

Find positive integers (a,b) so that sqrt(37 + 12*sqrt(7)) = a + b*sqrt(7). Answer in the form of "a, b".

Apr 3, 2022

#1
+118220
+1

$$\sqrt{37 + 12\sqrt7}= a + b\sqrt7\\ 37+12\sqrt7=a^2+7b^2+2ab\sqrt7\\ 12=2ab\\ ab=6\\ b=\frac{6}{a}\\ so\\ 37=a^2+7b^2\\ 37=a^2+\frac{7*36}{a^2}\\ 37a^2=a^4+252\\ 0=a^4-37a^2+252\\ let\;x=a^2\\ 0=x^2-37x+252\\ x=\frac{37\pm \sqrt{37^2-1008}}{2}\\ x=\frac{37\pm \sqrt{361}}{2}\\ x=\frac{37\pm 19}{2}\\ x=28 \;\;or \;\;x=8\\ a^2=28 \;\;or \;\;a^2=8\\ a=\pm 2\sqrt7 \;\;or \;\;a=\pm 2\sqrt2\\$$

You need to check all these answers, they will not all work some will need to be discarded.

You also need to check my working for careless errors

LaTex

\sqrt{37 + 12\sqrt7}= a + b\sqrt7\\
37+12\sqrt7=a^2+7b^2+2ab\sqrt7\\
12=2ab\\
ab=6\\
b=\frac{6}{a}\\
so\\
37=a^2+7b^2\\
37=a^2+\frac{7*36}{a^2}\\
37a^2=a^4+252\\
0=a^4-37a^2+252\\
let\;x=a^2\\
0=x^2-37x+252\\
x=\frac{37\pm \sqrt{37^2-1008}}{2}\\
x=\frac{37\pm \sqrt{361}}{2}\\
x=\frac{37\pm 19}{2}\\
x=28 \;\;or \;\;x=8\\
a^2=28 \;\;or \;\;a^2=8\\
a=\pm 2\sqrt7 \;\;or \;\;a=\pm 2\sqrt2\\

Apr 4, 2022