Gujarat BoardEnglish MediumSTD 11 ScienceMATHSQuadratic Equations3 Marks
Question
Show the following quadratic equation: $2\text{x}^2+\sqrt{15}\ \text{ix}-\text{i}=0$
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Answer
$2\text{x}^2+\sqrt{15}\ \text{ix}-\text{i}=0$ Comparing the given Equation with the general form $ax^2 + bx + c = 0$, we get a = 2, $\text{b}=\sqrt{15}\text{i},\text{c}=-\text{i}$ Substituting a and in. $\alpha=\frac{-\text{b}+\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$ and $\beta=\frac{-\text{b}-\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$ $\text{a}=\frac{-\sqrt{15}\text{ i}+\sqrt{-15+8}\text{ i}}{4}$ and $\beta=\frac{-\sqrt{15}\text{ i}-\sqrt{-15+8}\text{ i}}{4}$ Let $\sqrt{-15+8\text{i}}=\text{a}+\text{bi}$
$\Rightarrow-15+8 i=(a+b i)^2 \Rightarrow-15+8 i=a^2-b^2+2 a b i \Rightarrow a^2-b^2=-15 \text { and } 2 a b i=8 i $
$\text { Now }\left(a^2+b^2\right)=\left(a^2-b^2\right)+4 a^2 b^2=\left(a^2+b^2\right)=(15)^2+$$64=289 \Rightarrow a^2+b^2=17 \text { Solving } a^2-b^2=-15 \text { and } a^2+b^2=17 \text {, we get } a^2=1 \text { and } b^2=16$
$\Rightarrow\text{a}=\pm1$ and $\text{b}=\pm4$ ⇒ a = 1, b = 4 or a = -1, b = -4 $\therefore\sqrt{-15+8\text{i }}=1+4\text{i},-1-4\text{i}$ when $\sqrt{-15+8}\text{i}=1+4\text{i}$ $\alpha=\frac{-\sqrt{15}\text{i}+1+4\text{i}}{4}=\frac{1+(4-\sqrt{15})\text{i}}{4}$ and $\beta=\frac{-\sqrt{15}\text{i}-(1+4\text{i})}{4}=\frac{-1-(4+\sqrt{15})\text{i}}{4}$ When $\sqrt{-15+8\text{i}}=-1-4\text{i}$ $\alpha=\frac{-\sqrt{15}\text{i}-1-4\text{i}}{4}=\frac{-1-(4+\sqrt{15})}{4}$ and $\beta=\frac{-\sqrt{15}\text{i}-(-1-4\text{i})}{4}=\frac{1+(4-\sqrt{5})\text{i}}{4}$
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