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Atrend February 25th, 2019 10:46 AM

Inequality with square roots
 
Hello,

When practising, I was trying to solve following problem:
sqrt (x - 3) + sqrt (x + 5) > sqrt (x + 4)

The solution is apparently x > 3, but I get x > ((-12 + sqrt (912))/6 (which is 3.03322).Could anyone help me and give me some pointers as to how to tackle this problem please?

I start by actually squaring the numbers (condition: x > 3, otherwise the numbers in the square root will get negative), so I get:
x - 3 + 2(sqrt((x-3)(x+5))) + x + 5 > x + 4
Then I arrange the numbers:
2 * sqrt (x^2 + 2x - 15) > -x + 2
I now want to square again (no negative number under the square root under condition x > 3):
4(x^2 + 2x - 15) > x^2- 4x + 4
I work with this until I get:
3x^2 + 12x - 64 > 0.

Now, trying to factor the final one, I get D=144+768=912; with roots (-12+-sqrt(912))/6).
Taking the condition x>3 into account, my solutions seems to be: x>(-12+sqrt(912))/6.

Thanks in advance.

JeffM1 February 25th, 2019 11:56 AM

Quote:

Originally Posted by Atrend (Post 606257)
Hello,

When practising, I was trying to solve following problem:
sqrt (x - 3) + sqrt (x + 5) > sqrt (x + 4)

The solution is apparently x > 3, but I get x > ((-12 + sqrt (912))/6 (which is 3.03322).Could anyone help me and give me some pointers as to how to tackle this problem please?

I start by actually squaring the numbers (condition: x > 3, otherwise the numbers in the square root will get negative), so I get:
x - 3 + 2(sqrt((x-3)(x+5))) + x + 5 > x + 4
Then I arrange the numbers:
2 * sqrt (x^2 + 2x - 15) > -x + 2
I now want to square again (no negative number under the square root under condition x > 3):
4(x^2 + 2x - 15) > x^2- 4x + 4
I work with this until I get:
3x^2 + 12x - 64 > 0.

Now, trying to factor the final one, I get D=144+768=912; with roots (-12+-sqrt(912))/6).
Taking the condition x>3 into account, my solutions seems to be: x>(-12+sqrt(912))/6.

Thanks in advance.

$x < 3 \implies \sqrt{x - 3} \not \in \mathbb R.$

And if we are dealing with complex numbers, order is not relevant.

So, for the problem to make any sense, we start with a condition that

$x \ge 3 \implies \sqrt{x - 3} \ge 0.$

Furthermore,

$5 > 4 \implies x + 5 > x + 4 \text { for any value of } x.$

$\therefore x \ge 3 \implies \sqrt{x + 5} > \sqrt{x + 4} \implies \sqrt{x - 3} + \sqrt{x + 5}> 0 + \sqrt{x + 4} = \sqrt{x + 4}.$

When you start squaring things, you can get spurious results.

Denis February 25th, 2019 12:30 PM

Quote:

Originally Posted by Atrend (Post 606257)
sqrt (x - 3) + sqrt (x + 5) > sqrt (x + 4)

The solution is apparently x > 3 ....

If x=3 then:
sqrt(0) + sqrt(8) > sqrt(7)

skipjack February 25th, 2019 04:49 PM

If x = 3 when squaring the second time, one effectively has 0 > -1, and that shouldn't be squared to give 0 > 1.


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