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Quadratic Equations — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSQuadratic Equations3 Marks
Question
Show the following quadratic equation:
$2\text{x}^2+\sqrt{15}\ \text{ix}-\text{i}=0$
✓

Answer

$2\text{x}^2+\sqrt{15}\ \text{ix}-\text{i}=0$
Comparing the given Equation with the general form
ax2 + 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 $\sqrt{-15+8\text{i}}=\text{a}+\text{bi}$
⇒ -15 + 8i = (a + bi)2
⇒ -15 + 8i = a2 - b2 + 2abi
⇒ a2 - b2 = -15 and 2abi = 8i
Now (a2 + b2) = (a2 - b2) + 4a2b2
⇒ (a2 + b2) = (15)2 + 64 = 289
⇒ a2 + b2 = 17
Solving a2 - b2 = -15 and a2 + b2 = 17, we get
 a2 = 1 and b2 = 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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