Let $\alpha$ and $\beta$ be the roots of the equation x2
- x - 1 = 0.
If pk
= ${\left( \alpha \right)^k} + {\left( \beta \right)^k}$
, k $\ge$ 1, then which one
of the following statements is not true?
Solution
x<sup>2</sup>
- x - 1 = 0
<br><br>$\therefore$ $\alpha$<sup>2</sup>
- $\alpha$ - 1 = 0
<br><br>$\Rightarrow$ $\alpha$<sup>2</sup> = $\alpha$ + 1
<br><br>$\therefore$ $\alpha$<sup>3</sup> = $\alpha$<sup>2</sup> + $\alpha$
<br><br>= $\alpha$ + 1 + $\alpha$
<br><br>= 2$\alpha$ + 1
<br><br>Now $\alpha$<sup>4</sup> = 2$\alpha$<sup>2</sup> + $\alpha$
<br><br>= 2($\alpha$ + 1) + $\alpha$
<br><br>= 3$\alpha$ + 2
<br><br>Now $\alpha$<sup>5</sup> = 3$\alpha$<sup>2</sup> + 2$\alpha$
<br><br>= 3($\alpha$ + 1) + 2$\alpha$
<br><br>= 5$\alpha$ + 3
<br><br>Given p<sub>k</sub>
= ${\left( \alpha \right)^k} + {\left( \beta \right)^k}$
<br><br>$\therefore$ p<sub>5</sub> = ${\left( \alpha \right)^5} + {\left( \beta \right)^5}$
<br><br>= 5$\alpha$ + 3 + 5$\beta$ + 3
<br><br>= 5($\alpha$ + $\beta$) + 6
<br><br>= 5(1) + 6 [As $\alpha$ + $\beta$ = 1]
<br><br>= 11
<br><br>Now p<sub>2</sub> · p<sub>3</sub>
<br><br>= ($\alpha$<sup>2</sup> + $\beta$<sup>2</sup>).($\alpha$<sup>3</sup> + $\beta$<sup>3</sup>)
<br><br>= ( $\alpha$ + 1 + $\beta$ + 1)(2$\alpha$ + 1 + 2$\beta$ + 1)
<br><br>= ( $\alpha$ + $\beta$ + 2)(2($\alpha$ + $\beta$) + 2)
<br><br>= (1 + 2)(2 + 2) = 12
<br><br>$\therefore$ p<sub>5</sub> $\ne$ p<sub>2</sub> ·p<sub>3</sub>
<br><br>So option (D) is wrong.
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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