Question

Solve by using quadratic formula:

$8. \: 5{}^{2x - 1 } + 5 {}^{ x+ 1} = 250$
Ans: 2

Class 10

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Answer 1

Discriminant = b² - 4ac

=> (25)² - 4(5)(-1250)
=> 625 + 5000
=> 5625

Hence, applying quadratic equation,

$t = \frac{ - b + \sqrt{d} }{2a} \: \: \: \: or \: \: \: \: \: t= \frac{ - b - \sqrt{d} }{2a} \\ \\ \\ => t = \frac{ - 25 + \sqrt{5625} }{2(1)} \: \: \: or \: \: \: \: t= \frac{ - 25 - \sqrt{5625} }{2(1)} \\ \\ \\ = > t = \frac{ - 25 + 75}{2} \: \: \: \: \: or \: \: \: \: \: t= \frac{ - 25 - 75}{2} \\ \\ \\ => t= \frac{50}{2} \: \: \: or \: \: \: \: \: t = \frac{ - 100}{2} \\ \\ \\ \\ t = 25 \: \: \: or \: \: \: t= -50$




Neglect, t = -50


t = 25
5^x = 25
5^x = 5^2
x = 2

Answer 2

$Given : 5^{2x - 1} + 5^{x + 1} = 250$

$= \ \textgreater \ 5^{2x} * 5^{-1} + 5^{x} * 5^{1} = 250$

$= \ \textgreater \ ( 5^{x})^2 * 5^{-1} + 5^{x} * 5^{1} = 250$

$= \ \textgreater \ \frac{ (5^{x})^2 }{5} + 5^{x} * 5^1 = 250$

Let 5^x = y.

$= \ \textgreater \ \frac{y^2}{5} + 5y = 250$

$= \ \textgreater \ y^2 + 25y = 1250$

= > y^2 + 25y - 1250 = 0

(1) 

$= \ \textgreater \ x = \frac{-b + \sqrt{b^2 - 4ac} }{2a}$

$= \ \textgreater \ \frac{-(25) \sqrt{(25)^2 - 4(1)(-1250)} }{2}$

$= \ \textgreater \ \frac{-25 + \sqrt{625 + 5000} }{2}$

$= \ \textgreater \ \frac{-25 + \sqrt{5625} }{2}$

$= \ \textgreater \ \frac{-25 + 75}{2}$

$= \ \textgreater \ \frac{50}{2}$

= > 25.



(2) 

$= \ \textgreater \ \frac{-b - \sqrt{b^2 - 4ac} }{2a}$

$= \ \textgreater \ \frac{-b - \sqrt{(25)^2 - 4(1)(-1250} }{2}$

$= \ \textgreater \ \frac{-(25) - \sqrt{625 + 5000} }{2}$

$= \ \textgreater \ \frac{-25 - 75}{2}$

= > -100/2

= > -50.


Therefore the values are 25,-50.

Neglect -ve values. Then the value is y = 25.

Now,

= > 5^x = 25

= > 5^x = 5^2

= > x = 2.



Hope this helps!