Evaluation of expressions

If a2-3a+1=0, find

(i)a2+1/a2
(ii) a3+1/a3

Result

x1 =  7
x2 =  7
y1 =  18
y2 =  18

Solution:

 a2−3a+1=0 a2−3a+1=0  p=1;q=−3;r=1 D=q2−4pr=32−4⋅1⋅1=5 D>0  a1,2=−q±D2p=3±52 a1,2=1.5±1.1180339887499 a1=2.6180339887499 a2=0.38196601125011   Factored form of the equation:  (a−2.6180339887499)(a−0.38196601125011)=0 x1=a12+1/a12=2.6182+1/2.6182=7 \ \\ a^2-3a+1=0 \ \\ a^2 -3a +1=0 \ \\ \ \\ p=1; q=-3; r=1 \ \\ D=q^2 - 4pr=3^2 - 4\cdot 1 \cdot 1=5 \ \\ D>0 \ \\ \ \\ a_{1,2}=\dfrac{ -q \pm \sqrt{ D } }{ 2p }=\dfrac{ 3 \pm \sqrt{ 5 } }{ 2 } \ \\ a_{1,2}=1.5 \pm 1.1180339887499 \ \\ a_{1}=2.6180339887499 \ \\ a_{2}=0.38196601125011 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (a -2.6180339887499) (a -0.38196601125011)=0 \ \\ x_{1}=a_{1}^2+1/a_{1}^2=2.618^2+1/2.618^2=7

Checkout calculation with our calculator of quadratic equations.

x2=a22+1/a22=0.3822+1/0.3822=7x_{2}=a_{2}^2+1/a_{2}^2=0.382^2+1/0.382^2=7
y1=a13+1/a13=2.6183+1/2.6183=18y_{1}=a_{1}^3+1/a_{1}^3=2.618^3+1/2.618^3=18
y2=a23+1/a23=0.3823+1/0.3823=18y_{2}=a_{2}^3+1/a_{2}^3=0.382^3+1/0.382^3=18

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