mathematics-class-11-unit-05-chapter-03-quadratic-equations-l-3

<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Quadratic equation lecture 3</B></Success></p>
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<li>Let p, q be two real numbers such that $p \neq 0, p^3 \neq \pm q$. If $\alpha, \beta$ are non-zero complex numbers such that $\alpha+\beta=-p$ and $\alpha^3+\beta^3 = q$, then a quadratic equation having $\alpha / \beta$ and $\beta / \alpha$ as its solutions is :</li>
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<p>(1) $(p^3+q)x^2-(p^3+2q)x+(p^3+q) = 0$</p>
<p>(2) $(p^3+q)x^2-(p^3-2q)x+(p^3+q) = 0$</p>
<p>(3) $(p^3-q)x^2-(5p^3-2q)x+(p^3-q) = 0$</p>
<p>(4) $(p^3-q)x^2-(5p^3+2q)x+(p^3-q) = 0$</p>
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<p>$\frac{\alpha}{\beta}+\frac{\beta}{\alpha} =\frac{p^2-2 \frac{p^3+q}{3 p}}{\frac{p^3+q}{3 p}}=\frac{3 p^3-2 p^3-2 q}{(p^3+q)}$</p>
<p>$=\frac{p^3-2 q}{p^3+q}$</p>
<p>$x^2-(\frac{p^3-2 q}{p^3+q})x+1=0$</p>
<p>$(p^3+q)x^2-(p^3-2 q)x+p^3+q=0$</p>
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<p>$\alpha^2-6 \alpha-2=0 \quad \biggl[\alpha^2-2=6 \alpha\biggl]$</p>
<p>$\beta^2-6 \beta-2=0 \quad \biggl[\beta^2-2=6 \beta\biggl]$</p>
<p>$\frac{a_{10}-2a_8}{2a_9} =\frac{\alpha^{10}-\beta^{10}-2 \alpha^8+2 \beta^8}{2 \alpha^9-2 \beta^9}$</p>
<p>$=\frac{\alpha^8(\alpha^2-2)-\beta^8(\beta^2-2)}{2(\alpha^9-\beta^9)}$</p>
<p>$=\frac{6 \alpha^9-6 \beta^9}{2(\alpha^9-\beta^9)}=\frac{6}{2}=3$</p>
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<li>Let $\alpha, \beta$ be the solutions of $x^2-x-1=0$, with $\alpha>\beta$. Let $a_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}$ for $n \geq 1, b_1=1$ and $b_n={a_{n-1}+a_{n+1}}$ for $n \geq 2$. Then which of the following options is/are correct?</li>
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<p>(1) $a_1+a_2+\cdots+a_n=a_{n+2}-1$ for all $n \geq 1$</p>
<p>(2) $b_n=\alpha^n+\beta^n$ for all $n \geq 1$</p>
<p>(3) $\sum_{n\geq1} \frac{b_n}{10^n}=\frac{8}{89}$</p>
<p>(4) $\sum_{n\geq1} \frac{a_n}{10^n}=\frac{10}{89}$</p>
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<p>$\beta+2$</p>
<p>$\frac{1-\sqrt{5}}{2}+2$</p>
<p>$\frac{\sqrt{5}-\sqrt{5}}{2}$
$ =\sqrt{5}(\frac{\sqrt{5}-1}{2})$
$=-\sqrt{5} \beta$</p>
<p>$\alpha-\beta=\sqrt{5}$</p>
<p>$\alpha+\beta=1$</p>
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<p>$\sum_{n \geqslant 1} \frac{b_n}{10^n}=\sum_{n \geqslant 1} \frac{\alpha^n+\beta^n}{10^n}=\sum_{n \geqslant 1} \frac{\alpha^n}{10^n}+\sum_{n \geqslant 1} \frac{\beta^n}{10^n}$</p>
<p>$=\frac{\alpha / 10}{1-\alpha / 10}+\frac{\beta / 10}{1-\beta / 10}$</p>
<p>$=\frac{\alpha}{10-\alpha}+\frac{\beta}{10-\beta}$
$=\frac{\alpha(10-\beta)+\beta(10-\alpha)}{(10-\alpha)(10-\beta)}$</p>
<p>$=\frac{10(\alpha+\beta)-2 \alpha \beta}{100-10(\alpha+\beta)+\alpha \beta}$
$=\frac{10+2}{100-10-1}$
$=\frac{12}{89}$</p>
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<p>$\sum_{n \geqslant 1} \frac{a_n}{10^n}=\sum_{n \geqslant 1} \frac{\alpha^n-\beta^n}{(\alpha-\beta) 10^n}$</p>
<p>$=\frac{1}{\alpha-\beta} \sum_{n \geqslant 1} \frac{\alpha^n}{10^n}-\frac{1}{\alpha-\beta} \sum_{n \geqslant 1} \frac{\beta^n}{10^n}$</p>
<p>$=\frac{1}{\alpha-\beta}(\frac{\alpha}{10-\alpha}-\frac{\beta}{10-\beta})$
$=\frac{1}{\alpha-\beta} \cdot\{\frac{10 \alpha-\alpha \beta-10 \beta+\alpha \beta}{100-10(\alpha+\beta)+\alpha \beta}\}$</p>
<p>$ =\frac{1}{\alpha-\beta} \cdot \frac{10(\alpha-\beta)}{100-10(\alpha+\beta)+\alpha \beta}$
$=\frac{10}{100-10-1}$
$=\frac{10}{89}$</p>
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<p>$b_n =a_{n-1}+a_{n+1}$</p>
<p>$=\frac{\alpha^{n-1}-\beta^{n-1}}{\alpha-\beta}+\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}$</p>
<p>$ =\frac{\alpha^{n-1}(\alpha^2+1)-\beta^{n-1}(\beta^2+1)}{\alpha-\beta}$
$ =\frac{\alpha^{n-1}(\alpha+2)-\beta^{n-1}(\beta+2)}{\alpha-\beta}$</p>
<p>$=\frac{\alpha^n-\beta^n}{\alpha-\beta}+2 \frac{\alpha^{n-1}-\beta^{n-1}}{\alpha-\beta}$</p>
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<p>$=1-b+b^2-b^3$</p>
<p>$b^3+3b=0$</p>
<p>$b(b^2+3)=0$</p>
<p>$b=0 \quad or \quad b^2+3=0$</p>
<p>$b=0\qquad$, $b^2=-3\quad$(Not in the options)</p>
<p>$b= \pm \sqrt{3} i$</p>
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