Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 207897 by hardmath last updated on 29/May/24

$$\mathrm{Find}:$$$$\sqrt{12\cdot 13\cdot 14\cdot 15+1}=?$$

Answered by Frix last updated on 29/May/24

$$\sqrt{x(x+1)(x+2)(x+3)+1}=$$$$=\sqrt{x^4+6x^3+11x^2+6x+1}=$$$$=\sqrt{{(x^2+3x+1)}^2}=$$$$=x^2+3x+1\stackrel{x=12}{=}181$$

Answered by Rasheed.Sindhi last updated on 30/May/24

$$letx=13\frac{1}{2}$$$$=\sqrt{(x-\frac{3}{2})(x+\frac{3}{2})(x-\frac{1}{2})(x+\frac{1}{2})+1}$$$$=\sqrt{(x^2-\frac{9}{4})(x^2-\frac{1}{4})+1}$$$$=\sqrt{x^4-\frac{x^2}{4}-\frac{9x^2}{4}+\frac{9}{16}+1}$$$$=\sqrt{x^4-\frac{5x^2}{2}+\frac{25}{16}}$$$$=\sqrt{{\left(x^2\right)}^2-2\left(x^2\right)\left(\frac{5}{4}\right)+{\left(\frac{5}{4}\right)}^2}$$$$=\sqrt{{(x^2-\frac{5}{4})}^2}$$$$=\mid x^2-\frac{5}{4}\mid$$$$\Rightarrow \mid {\left(\frac{27}{2}\right)}^2-\frac{5}{4}\mid$$$$=\frac{{27}^2-5}{4}=181$$

Answered by BaliramKumar last updated on 31/May/24

$$\sqrt{a\cdot (a+d)\cdot (a+2d)\cdot (a+3d)+d^4}=a(a+3d)+d^2$$$$a=12,\mathrm{d}=1$$$$12(12+3\times 1)+1^2={12}^2+12\times 3+1$$$$144+36+1=181$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com