Warm up problem 14: DONT BASH

NO, the trick is not guess and check... If I had made the numbers uglier, it wouldn't work.

But anyway, while in my Alg class, again, I figured out a way to substitute when problems had two square roots. By substituting a for the radical with the lower x coefficient, we can eliminate one radical.

Set $a=\sqrt{x-2}$

$a+\sqrt{11x}=14$

Now, to eliminate the second radical, we match the x-coefficients of a and the radical and subtract the remaining constant.

$a+\sqrt{11a^2+22}=14$

$\sqrt{11a^2+22}=14-a$

$11a^2+22=a^2-28a+196$

$10a^2+28a-174=0$

$2(5a^2+14a-87)=0$

$(5a+29)(a-3)=0$

Since a represents a radical, it can't be negative and thus $5a-29$ is extraneous.

$a-3=0\Rightarrow a=3$

Resubstituting $\sqrt{x-2}=a$

$\sqrt{x-2}=3$

$\boxed{x=11}$

Given

$\sqrt{x-2}+\sqrt{11x}=14$

Checking the domain of two square-root values , we can say $x \geq 2$

And also we know, both square-root values must be square-root free after putting the value of x

If for making $\sqrt{11x}$ square-root free, we will put the positive multiples of 11

Putting x=11 satisfies the given equation

And also if we put 22, 33 ...and so on, then $\sqrt{11x}$ will be greater than 14. So we will neglect taking these values.

Therefore $\large \boxed{x=11}$

enjoy!

$\sqrt{x-2}+\sqrt{11x}=14\longrightarrow 12x-2+2\sqrt{11x^2 -22x}=14^2$ $14(7)-6x+1=\sqrt{11x^2 -22x}\longrightarrow (99-6x)^2=11x^2 -22x$ $9801-1188x+36x^2=11x^2-22x\longrightarrow 25x^2-1166x+9801=0$ $x=\dfrac{1166\pm \sqrt{1166^2 -2^2\times 99^2 \times 5^2}}{50}$ $x=\dfrac{1166\pm \sqrt{1166^2 -(2\times 99 \times 5)^2}}{50}$ $x=\dfrac{1166\pm \sqrt{1166^2 -990^2}}{50}$ $x=\dfrac{1166\pm \sqrt{(1166+990)(1166-990}}{50}$ $x=\dfrac{1166\pm \sqrt{2156\times 176}}{50}$ $x=\dfrac{1166\pm \sqrt{2156\times 16 \times 11}}{50}$ $x=\dfrac{1166\pm \sqrt{196 \times16\times 121}}{50}$ $x=\dfrac{1166\pm (14\times 4 \times 11)}{50}$ $x=\dfrac{22(53\pm 28)}{50}$ $x=\dfrac{891}{25}~(extraneous), ~~ \boxed {11}$

Since 14 is a whole number, the result of every root had to be a whole numbers as well. The easiest way (not recurring to extensive algebra) was, in my opinion, to guess by breaking down the second root multiplying the x by 11. This number (11) also worked (luckily) for the first root, and the resulting sum was exactly 14.

The root of 11x can be easily get by putting 11 inplace of x..so I tried putting 11 in place of x....and this way I got the answer

Using only whole squares 11*11sqrt=11 11-2sqrt=3 3+11=14

clearly you have match 14 = 14 so just find a number which breaks the square-root( means which makes square inside the square-root ). So to break square-root(11 x) is 11 because square-root(11 11) = 11 and square-root(11-2)=3 which sums to 14 so that required number is 11