The root-s- of 15-x2 - 4- ndash- 2-x - 2- - 1 is -are-a- -5 and 3- -b- plusmn- 2- -c- 2 only- -d- -3 and 5- -e- 3 only

Dear student 15x2 4 2x 2=1 #8658;15x 2x+2 2x 2=1 #8658;15 2x+2x 2x+2=1 #8658;15 2x 4x 2x+2=1 #8658;11 2x=x2 4 #8658;x2 4 11+2x=0 #8658;x2+2x 15=0 ...

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Byju's Answer

Standard XII

Mathematics

Monotonically Decreasing Functions

The root(s) o...

Question

The root(s) of 15/(x² - 4) – 2/(x - 2) = 1 is (are):
(a) -5 and 3, (b) ± 2, (c) 2 only, (d) -3 and 5, (e) 3 only

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Solution

Dear student
$\frac{15}{(x^{2} - 4)} - \frac{2}{(x - 2)} = 1 \Rightarrow \frac{15}{(x - 2) (x + 2)} - \frac{2}{x - 2} = 1 \Rightarrow \frac{15 - 2 (x + 2)}{(x - 2) (x + 2)} = 1 \Rightarrow \frac{15 - 2 x - 4}{(x - 2) (x + 2)} = 1 \Rightarrow 11 - 2 x = x^{2} - 4 \Rightarrow x^{2} - 4 - 11 + 2 x = 0 \Rightarrow x^{2} + 2 x - 15 = 0 \Rightarrow x^{2} + 5 x - 3 x - 15 = 0 \Rightarrow x (x + 5) - 3 (x + 5) = 0 \Rightarrow (x - 3) (x + 5) = 0 \Rightarrow x = 3 o r x = - 5$
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The root(s) of 15/(x2 - 4) – 2/(x - 2) = 1 is (are): (a) -5 and 3, (b) ± 2, (c) 2 only, (d) -3 and 5, (e) 3 only

Dear student 15x2-4-2x-2=1⇒15x-2x+2-2x-2=1⇒15-2x+2x-2x+2=1⇒15-2x-4x-2x+2=1⇒11-2x=x2-4⇒x2-4-11+2x=0⇒x2+2x-15=0⇒x2+5x-3x-15=0⇒xx+5-3x+5=0⇒x-3x+5=0⇒x=3 or x=-5 Regards

The root(s) of 15/(x² - 4) – 2/(x - 2) = 1 is (are):
(a) -5 and 3, (b) ± 2, (c) 2 only, (d) -3 and 5, (e) 3 only