Roots of Roots
Find the value of x satisfying the equation 3 x + 5 − x − 2 = 3 .
Details and assumptions
If the square root signs are not displaying clearly for you, the equation is equivalent to
[ 3 ( x + 5 ) 2 1 − ( x − 2 ) 2 1 ] 2 1 = 3 .
The answer is 11.
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6 solutions
same
I used rather a brute force way
Remove the sqrt from the l.h.s
3*sqrt(x+5) - sqrt(x-2) = 9
3*sqrt(x+5) = 9 + sqrt(x-2) Rearranging
(3)^2 * (x+5) = (9 + sqrt(x-2))^2 Eliminate square root from the l.h.s term
9x + 45 = 81 + x - 2 + 18*sqrt(x-2)
8x -34 = 18*sqrt(x-2) Simplifying
4x - 17 = 9*sqrt(x-2)
(4x - 17)^2 = (9)^2 * (x-2) Eliminate square root from the r.h.s term
16x^2 -136x + 289 = 81(x-2)
16x^2 - 217x + 451 = 0 Rearrange
By finding the roots of quadratic equation will give us
x = 11 and x = 2.5625
Since ans is an integer, x = 11
please show the solution of last quadratic eqatios roots
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You can use the formula
x = 2 a − b ± b 2 − 4 a c
for the values of a, b, and c,
a x 2 + b x + c = 0
Hence,
a = 1 6
b = − 2 1 7
c = 4 5 1
Substituting these values into the given formula will give us,
x = 2 ( 1 6 ) − ( − 2 1 7 ) ± ( − 2 1 7 ) 2 − 4 ( 1 6 ) ( 4 5 1 )
x = 3 2 2 1 7 ± 1 8 2 2 5
Hence, the two roots/solutions of x in the quadratic equation are:
x = 3 2 2 1 7 + 1 8 2 2 5
and
x = 3 2 2 1 7 − 1 8 2 2 5
where the first solution should give us an integer which is 11
There is a lazier method, which is to directly factorise the quadratic equation. It should look like this
( x − 1 1 ) ( x + 2 . 5 6 2 5 ) = 0
x − 1 1 = 0 x = 1 1
This one I ended up solving analytically, and I would welcome a -much- better solution if anyone has it. But, for now, I'll just explain how I got it.
So, first and foremost, I got rid of the first square root on the left side by simply squaring - leaving us with 3*sqrt(x+5) - sqrt(x-2) = 9
At this point, I attempted to see if I could get a straightforward solution, but it became very complicated very quickly, and I decided to instead give my hand at solving it analytically.
First off, we know that X is -not- going to be a big number. 3*sqrt(x+5) is going to grow much faster than sqrt(x-2), so once we start getting into bigger numbers, that sqrt(x-2) will not be able to subtract our first term to form a number as small as 9.
So, I started by restricting myself to small numbers - something definitely under 20 by my guesstimation. I then made another logical leap - in order to get a nice, round 9, the X that we use has to lead to both square-roots coming out to whole numbers. I figured it fairly unlikely that we could have the two square roots arranged in just such a way that their decimals cancel out to form a whole 9, so it was a fairly good assumption. If you comb through the numbers that will make both of those square roots a whole number under 20, the first one I found was 11 - which is our answer.
3 sqrt(11 +5) - sqrt(11-2) = 3 sqrt(16) - sqrt(9) = 3*4 - 3 = 12 - 3 = 9
Algebraic solution is fairly simple, even simpler: after you get rid of the first square root on the left side, you square both sides of the equation again. The rest is just more algebra.
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That's what I did initially, actually. I don't tend to like getting too deep into tedious algebra, as this seemed to be, if it can be avoided, hence why I doubled back and did the analytic solution instead. I was hoping someone might have a nice, clean algebraic solution.
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I wish someone does. The algebra here wasn't that tedious to me, though. I usually go for the solution first and then look for prettier methods. If there actually exists a simpler way of doing this one, I'd like to know it. It's too late now, and I'm too lazy.
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@Sasha Brenner Socas – I'll give the algebra another go in a little while, see if I can come up with an elegant solution and post it here.
It can be noticed that because the right side of the equation is an integer, x + 5 and x − 2 must also (likely) be. This implies that the square roots must be of two integers which are perfect squares exactly 7 apart. The only two positive integers that fit these criteria are 9 and 1 6 - therefore, x must equal 1 1 to make both square roots have integer values. Plugging in 1 1 , we find that the equation is, in fact, equal to 3 .
squaring both d sides we get : 3(x+5)1/2 - (x-2)1/2 = 9
by looking at above equation, i noticed that it must b like 3*4 - 3. so if we take x=11, it wil satisfy the equation. :)
I solve this problem with trial and error, but I can explain how I worked. (1) Square both sides such that 3\sqrt(x+5)-\sqrt(x-2)=9 (2) Since 9 is multiple of 3, then 3\sqrt(x+5)-\sqrt(x-2) must also be the multiple of 3 (3) It is clear that 3.4-3 = 9, so I assumed that \sqrt(x+5)=4 and \sqrt(x-2)=1. Unpredictably, I got x=11. And also, I could simply check this value
after the squaring it is clear that both the terms must be integers and then hit and trail will easily if we smartly eliminates the possibilities.
We can see that x + 5 and x − 2 have a difference of 7. Since 9 and 16 are the only perfect squares with a difference of 7 and x + 5 > x − 2 , solving the equations x + 5 = 1 6 and x − 2 = 9 gives us 11.