First slide

Theory of equations

Question

The roots of the equation $(a + \sqrt{b})^{x^{2} - 15} + (a - \sqrt{b})^{x^{2} - 15} = 2 a$ , where $a^{2} - b = 1$ , are

Moderate

A

$\pm 2, \pm \sqrt{3}$
B

$\pm 4, \pm \sqrt{14}$
C

$\pm 3, \pm \sqrt{5}$
D

$\pm 6, \pm \sqrt{20}$

Solution

We have,
$a - \sqrt{b} = \frac{(a - \sqrt{b}) (a + \sqrt{b})}{a + \sqrt{b}} = \frac{a^{2} - b}{a + \sqrt{b}} = \frac{1}{a + \sqrt{b}} [∵ a^{2} - b = 1]$
So, by putting $(a + \sqrt{b})^{x^{2} - 15} = y$ , the given equation becomes
$\begin{aligned} y + \frac{1}{y} = 2 a \\ \Rightarrow & y^{2} - 2 ay + 1 = 0 \\ \Rightarrow & (y - a)^{2} = a^{2} - 1 \\ \Rightarrow & y - a = \pm \sqrt{a^{2} - 1} \\ \Rightarrow & y - a = \pm \sqrt{b} \\ \Rightarrow & y = a \pm \sqrt{b} \\ \Rightarrow & (a + \sqrt{b})^{x^{2} - 15} = a + \sqrt{b}, a - \sqrt{b} \\ \Rightarrow & x^{2} - 15 = 1 or, x^{2} - 15 = - 1 \Rightarrow x = \pm 4, x = \pm \sqrt{14} . \end{aligned}$

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