# If $$\dfrac{(x+2)!}{(2x-1)!}\times\dfrac{(2x+1)!}{(x+3)!}=\dfrac{72}{7}$$, where x ∈ N then find the value of x?

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DSSSB Patwari Question Paper 16th June 2019 Shift 1
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1. $$\dfrac{-54}{28}$$
2. 4
3. -4
4. $$\dfrac{54}{28}$$

Option 2 : 4
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DSSSB Patwari Mental Ability Sectional Test 1
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## Detailed Solution

$$\frac{{(x + 2)!}}{{(2x - 1)!}} \times \frac{{(2x + 1)!}}{{(x + 3)!}} = \frac{{72}}{7}$$

$$\frac{{(x + 2)!}}{{(2x - 1)!}} \times$$$$\frac{{(2x + 1)(2x)(2x - 1)!}}{{(x + 3)(x + 2)!}} = \frac{{72}}{7}$$

$$= \frac{{2x(2x + 1)}}{{x + 3}} = \frac{{72}}{7}$$

$$= \frac{{x(2x + 1)}}{{x + 3}} = \frac{{36}}{7}$$

= 14x2 + 7x = 36x + 108

= 14x2 - 29x - 108 = 0

$$x = \frac{{ - ( - 29) \pm \sqrt {{{( - 29)}^2} + 4 \times 14 \times 108} }}{{2 \times 14}}$$

$$= \frac{{29 \pm 83}}{{28}}$$

$$x = \frac{{29 + 83}}{{28}}$$$$x = \frac{{29 - 83}}{{28}}$$

x = 4, -1.92

Given x ϵ N, N → Natural number

x = 4
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