Algebra Problems

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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

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ilikemath247365

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ilikemath247365

#1 h Apr 14, 2025, 4:52 PM

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Find all real $(a, b)$ with $a + b = 1$ such that

$(a + \frac{1}{a})^{2} + (b + \frac{1}{b})^{2} = \frac{25}{2}$ .

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#2 h Apr 14, 2025, 4:57 PM

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xiooix !
a^2+b^2+1/a^2+1/b^2=17/2
let x=a^2 and y=b^2

thus x+y+1/x+1/y=17/2 so (x+y)(1+1/xy)=17/2

thus xy=2/15 and x+y=1, and the rest is computation.
oops wrong

This post has been edited 1 time. Last edited by fruitmonster97, Apr 14, 2025, 5:25 PM

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ilikemath247365

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ilikemath247365

#3 h Apr 14, 2025, 5:15 PM

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I think that is incorrect. My solution:

Click to reveal hidden text

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ilikemath247365

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ilikemath247365

#5 h Apr 14, 2025, 5:22 PM

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Wait sorry never mind I didn't see your substution of x = a^2 and y = b^2.

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#6 h Apr 14, 2025, 5:24 PM

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ilikemath247365 wrote:

I think that is incorrect. My solution:

Click to reveal hidden text

oops im sobad we have a+b=1 not a^2+b^2=1

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ilikemath247365

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ilikemath247365

#7 h Apr 14, 2025, 5:25 PM

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Oh yeah.

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#8 h Apr 14, 2025, 6:27 PM

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$2\left( \left( a + \frac{1}{a} \right)^2 + \left(b + \frac{1}{b} \right)^2 \right) \geq \left( a + \frac{1}{a} + b + \frac{1}{b} \right)^2$ $RHS = a + \frac{a+b}{a} + b + \frac{a + b}{b} = 3 + \frac{a}{b} + \frac{b}{a} \geq 5,$ $QED.$

edit: nvm i didnt see this wasnt a proof. oh well, the rest is easy anyway. equality must occur at $a + \frac{1}{a} = b +\frac{1}{b},$ or $a=b.$

This post has been edited 1 time. Last edited by anduran, Apr 14, 2025, 6:29 PM

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#9 h Apr 14, 2025, 9:56 PM

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We have $a^2+b^2+{{a^2+b^2}\over {a^2b^2}}=17/2$ and with $p:=ab$ we get $1-2p+{{1-2p}\over {p^2}}=17/2$ hence $4p^3+15p^2+4p-2=0$ . Factoring gives $(4p-1)(p^2+4p+2)=0$ .
Now $p=a(1-a)\le 1/4$ hence each value from $p=1/4, -2\pm \sqrt{2}$ gives a solution.

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lgx57

#10 h Apr 18, 2025, 1:28 PM

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ilikemath247365 wrote:

Find all real $(a, b)$ with $a + b = 1$ such that

$(a + \frac{1}{a})^{2} + (b + \frac{1}{b})^{2} = \frac{25}{2}$ .

Only for $a,b>0$ .

$(a + \frac{1}{a})^{2} + (b + \frac{1}{b})^{2} = \frac{25}{2} \Leftrightarrow (a^2+b^2)+(\frac{1}{a^2}+\frac{1}{b^2})=\frac{17}{2}$

$(a^2+b^2)+(\frac{1}{a^2}+\frac{1}{b^2}) \ge 2ab+\frac{2}{ab}$ . Because $a,b>0$ , $ab \in[0,\frac{1}{4}]$ .

Obviously , $f(x)=2x+\frac{2}{x}$ is monotonic decline in $[0,\frac{1}{4}]$ .

So $2ab+\frac{2}{ab} \ge f(\frac{1}{4})=\frac{17}{4}$

Then $(a + \frac{1}{a})^{2} + (b + \frac{1}{b})^{2} \ge \frac{17}{4} +2 = \frac{25}{2}$

The equality holds iff $a=b$ and $ab=\frac{1}{4}$ .

So the answer is $(a,b)=(\frac{1}{2},\frac{1}{2})$ .

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SomeonecoolLovesMaths

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SomeonecoolLovesMaths

#11 h Apr 18, 2025, 2:14 PM • 1 Y

Y by EntangledElectron99

If $\mid z -2 + i \mid \leq 2$ , find the greatest and least values of $\mid z \mid$ .

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lgx57

#12 h Apr 19, 2025, 4:25 AM

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SomeonecoolLovesMaths wrote:

If $\mid z -2 + i \mid \leq 2$ , find the greatest and least values of $\mid z \mid$ .

You can solve this problem in complex plane.

$\mid z-2+i \mid$ means distance from $(2,-1)$ to $z$ , so $\mid z-i+i \mid \leq 2$ stands for a circle with a center at $(2,-1)$ and a radius of $2$ .

So $\mid z \mid \in [\sqrt{5}-2,\sqrt{5}+2]$

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