Form the equation

Algebra Level 3

Which of the following equations whose roots are 3 + 5 i 2 7 \dfrac{3 + 5i\sqrt{2}}{7} and 3 − 5 i 2 7 \dfrac{3 - 5i\sqrt{2}}{7} ?

49 x 2 + 42 x − 41 = 0 49x^2 + 42x - 41 = 0 49 x 2 + 42 x + 59 = 0 49x^2 + 42x + 59 = 0 49 x 2 − 42 x + 59 = 0 49x^2 - 42x + 59 = 0 49 x 2 − 42 x − 41 = 0 49x^2 - 42x - 41 = 0

Munem Shahriar by Munem Shahriar , 18, Bangladesh

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2 solutions

Ravneet Singh
Jul 3, 2017

Let x = 3 ± 5 i 2 7 x = \dfrac{3 \pm 5i\sqrt{2}}{7}

7 x = 3 ± 5 i 2 7x = 3 \pm 5i\sqrt{2}

7 x − 3 = ± 5 i 2 7x - 3 = \pm 5i\sqrt{2}

Squaring both sides

( 7 x − 3 ) 2 = ( ± 5 i 2 ) 2 (7x - 3)^2 = (\pm 5i\sqrt{2})^2

49 x 2 − 42 x + 9 = − 50 49x^2 - 42x + 9 = - 50

49 x 2 − 42 x + 59 = 0 49x^2 - 42x + 59 = 0

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Let α = 3 + 5 2 i 7 \alpha = \dfrac {3+5\sqrt2i}7 , then α ˉ = 3 − 5 2 i 7 \bar{\alpha} = \dfrac {3-5\sqrt2i}7 . By Vieta's formula , the equation is:

x 2 − ( α + α ˉ ) x + α α ˉ = 0 x 2 − ( 3 + 5 2 i 7 + 3 − 5 2 i 7 ) x + 3 + 5 2 i 7 × 3 − 5 2 i 7 = 0 x 2 − 6 7 x + 9 + 50 49 = 0 ⟹ x 2 − 42 x + 59 = 0 \begin{aligned} x^2 - (\alpha + \bar{\alpha})x + \alpha \bar{\alpha} & = 0 \\ x^2 - \left(\frac {3+5\sqrt2i}7 + \frac {3-5\sqrt2i}7\right) x + \frac {3+5\sqrt2i}7 \times \frac {3-5\sqrt2i}7 & = 0 \\ x^2 - \frac 67 x + \frac {9+50}{49} & = 0 \\ \implies x^2 - 42x + 59 & = 0 \end{aligned}

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