The sum of the real roots of the equation 7-4-3x2-8-7-4-3x2-8-14 is

(7+4√(3))^x^2 8+(7 4√(3))^x^2 8=14 (7+4√(3))×(7 4√(3)) = 49 16× 3 = 1 ⇒ 7 4√(3)=17+4√(3) Assuming nbsp;(7+4√(3))^x^2 8 = t, we get t+ 1 t = 14 ⇒ t^2 14t ...

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Standard XII

Mathematics

Methods of Solving a Quadratic Equation

The sum of th...

Question

The sum of the real roots of the equation $(7 + 4 \sqrt{3})^{x^{2} - 8} + (7 - 4 \sqrt{3})^{x^{2} - 8} = 14$ is

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Solution

$(7 + 4 \sqrt{3})^{x^{2} - 8} + (7 - 4 \sqrt{3})^{x^{2} - 8} = 14$

$(7 + 4 \sqrt{3}) \times (7 - 4 \sqrt{3}) = 49 - 16 \times 3 = 1$
$\Rightarrow 7 - 4 \sqrt{3} = \frac{1}{7 + 4 \sqrt{3}}$
Assuming $(7 + 4 \sqrt{3})^{x^{2} - 8} = t$ , we get
$t + \frac{1}{t} = 14 \Rightarrow t^{2} - 14 t + 1 = 0 \Rightarrow t = \frac{14 \pm \sqrt{196 - 4}}{2} \Rightarrow t = 7 \pm 4 \sqrt{3}$

Now, $(7 + 4 \sqrt{3})^{x^{2} - 8} = 7 \pm 4 \sqrt{3}$
$\Rightarrow x^{2} - 8 = \pm 1 \Rightarrow x^{2} = 7, 9 \Rightarrow x = \pm \sqrt{7}, \pm 3$

Hence, the sum of real roots $= 0$

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The sum of the real roots of the equation (7+4√3)x2−8+(7−4√3)x2−8=14 is

(7+4√3)x2−8+(7−4√3)x2−8=14 (7+4√3)×(7−4√3)=49−16×3=1 ⇒7−4√3=17+4√3 Assuming (7+4√3)x2−8=t, we get t+1t=14⇒t2−14t+1=0⇒t=14±√196−42⇒t=7±4√3 Now, (7+4√3)x2−8=7±4√3 ⇒x2−8=±1⇒x2=7,9⇒x=±√7,±3 Hence, the sum of real roots =0

The sum of the real roots of the equation $(7 + 4 \sqrt{3})^{x^{2} - 8} + (7 - 4 \sqrt{3})^{x^{2} - 8} = 14$ is