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Let r and s be the roots of 3x^2+4x+12 = x^2 - 5x + 7. Find r^2+s^2.

 Apr 21, 2021
 #1
avatar+544 
0

3x2 + 4x + 12 = x2 - 5x + 7

2x2 + 9x + 5 = 0

x2 + \(\frac{9}{2}\)x + \(\frac{5}{2}\) = 0

x2 + \(\frac{9}{2}\)x + \(\frac{81}{16}\) - \(\frac{81}{16}\) + \(\frac{5}{2}\) = 0

(x + \(\frac{9}{4}\) )2 - \(\frac{81}{16}\)\(\frac{40}{16}\) = 0

(x + \(\frac{9}{4}\) )2 = \(\frac{41}{16}\)

x + \(\frac{9}{4}\) = ± \(\frac{\sqrt{41}}{4}\)

x = ± \(\frac{\sqrt{41} - 9}{4}\)

 

r = \(\frac{\sqrt{41} - 9}{4}\)

s = \(\frac{9 - \sqrt{41}}{4}\)

 

r2 + s2 = \((\frac{\sqrt{41} - 9}{4})^2 + (\frac{9 - \sqrt{41}}{4})^2\)

\(\frac{(\sqrt{41} - 9)^2 + (9 - \sqrt{41})^2}{16}\)

\(\frac{122 - 18\sqrt{41} + 122 - 18\sqrt{41}}{16}\)

\(\frac{2(122 - 18\sqrt{41})}{16}\)

\(\frac{122 - 18\sqrt{41}}{8}\)

\(\frac{61 - 9\sqrt{41}}{4}\)

 

tell me if I did anything wrong :)

 Apr 21, 2021
 #2
avatar+420 
+1

When you get to 2x^2 + 9x + 5 = 0, you could just use Vieta's formulas. The sum \(r+s=-\frac{b}{a}=-\frac{9}{2}\), which means that \((r+s)^2=r^2+2rs+s^2=(-\frac{9}{2})^2=\frac{81}{4}\). The product \(rs=\frac{c}{a}=\frac{5}{2}\), which means \(2rs=5\). Notice that \(r^2+2rs+s^2-2rs=r^2+s^2=\frac{81}{4}-\frac{20}{4}=\boxed{\frac{61}{4}}\)

textot  Apr 22, 2021

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