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Algebra

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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
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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
+420
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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