Inter Exchangeable Symbols

Algebra Level 4

Let a a and b b be distinct numbers such that

  • b b is a root of the equation x 2 + a x + 10 = 0 x^2 + ax + 10 = 0 , and
  • a a is a root of the equation x 2 + b x + 10 = 0 x^2 + bx + 10 = 0 .

What is the relationship between a a and b b ?

a + b = 0 a + b= 0 a + b = 1 a + b= 1 a + b = 2 a +b = 2 a + b = 3 a+b=3 No such a a and b b exist

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5 solutions

Yatin Khanna
Jul 30, 2016

Since, a a is the root of x 2 + b x + 10 = 0 x^2+bx+10=0 ;
a 2 + a b + 10 = 0 a^2+ab+10=0 ................................. ( 1 ) (1)
Also, b b is the root of x 2 + a x + 10 = 0 x^2+ax+10=0 ;
b 2 + a b + 10 = 0 b^2+ab+10=0 ................................. ( 2 ) (2)
SUBTRACTING ( 2 ) (2) FROM ( 1 ) (1) ; we get:
a 2 b 2 = 0 a^2-b^2 =0
=> ( a + b ) ( a b ) = 0 (a+b)(a-b)=0
Now, since a a and b b are distinct;
a b 0 a-b \neq 0
Hence; ( a + b ) (a+b) should be 0 0 ; but since a ( a + b ) + 10 = 0 a(a+b) + 10 = 0 ; hence ( a + b ) 0 (a+b) \neq 0
So, No such a and b exist.
All credits to @Abhay Kumar

Let b , d b,d be the roots of x 2 + a x + 10 = 0 x^{2} + ax + 10 = 0 and a , c a,c the roots of x 2 + b x + 10 = 0 x^{2} + bx + 10 = 0 . Then by Vieta's

  • (i) b + d = a b + d = -a and b d = 10 bd = 10 , implying that b + 10 b = a b 2 + 10 = a b b + \dfrac{10}{b} = -a \Longrightarrow b^{2} + 10 = -ab , and

  • (ii) a + c = b a + c = -b and a c = 10 ac = 10 , implying that a + 10 a = b a 2 + 10 = a b a + \dfrac{10}{a} = -b \Longrightarrow a^{2} + 10 = -ab .

Comparing the final results of (i) and (ii) gives us that a 2 = b 2 a 2 b 2 = ( a b ) ( a + b ) = 0 a^{2} = b^{2} \Longrightarrow a^{2} - b^{2} = (a - b)(a + b) = 0 .

As a a and b b must be distinct we cannot have a b = 0 a - b = 0 , so we must have that a + b = 0 \boxed{a + b = 0} .

Edit: As Abhay Kumar has pointed out in the comments section, (and Jon Haussmann in the reports section), since a a is a root of x 2 + b x + 10 x^{2} + bx + 10 we should have that a 2 + b a + 10 = a ( a + b ) + 10 = 0 a^{2} + ba + 10 = a(a + b) + 10 = 0 . But if a + b = 0 a + b = 0 we would end up with a 0 + 10 = 0 a*0 + 10 = 0 , which is a falsehood, and thus the correct answer is that no such a a and b b exist.

Sir, As a a is root of x 2 + b x + 10 = 0 x^2+bx+10=0 .
Then, a 2 + a b + 10 = 0 a ( a + b ) 0 + 10 0 a^2+ab+10=0 \ \implies a \underbrace{(a+b)}_{0}+10 \not= 0 .

\therefore No such a a and b b exist.

A Former Brilliant Member - 4 years, 10 months ago

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Ah, yes, I should have noticed that. I see that the option "No such a a and b b exist" has been added to account for this fact.

Brian Charlesworth - 4 years, 10 months ago

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Okay.No problem! Please change your solution according to that. :)

A Former Brilliant Member - 4 years, 10 months ago

Nice observation, sir!

Yatin Khanna - 4 years, 10 months ago

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Lol!...I am just 16 presently...No need to say sir. :p

A Former Brilliant Member - 4 years, 10 months ago

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@A Former Brilliant Member Okay, sir. xD ;p :p

Yatin Khanna - 4 years, 10 months ago
Thomas Jacob
Aug 18, 2016

Since a is a root of the equation x 2 + b x + 10 x^2 +bx + 10 , we can rewrite it as a 2 + a b + 10 = 0 a^2 +ab + 10 = 0 Similarly, the second equation implies that b 2 + a b + 10 = 0 b^2 + ab + 10 = 0 By simply adding the two equations we get, a 2 + b 2 + 2 a b = 20 a^2 + b^2 + 2ab = -20 Which is nothing but ( a + b ) 2 = 20 (a+b)^2 = -20 But since a squared number cannot be negative, there exist no such a and b.

Jakub Bober
Aug 14, 2016

The is a solution is much more simple then the ones below. x 2 + a x + 10 = 0 x^{2}+ax+10=0 and x 2 + b x + 10 = 0 x^{2}+bx+10=0 , so x 2 + a x + 1 = x 2 + b x + 10 x^{2}+ax+1=x^{2}+bx+10 . Substracting x 2 + 10 x^{2}+10 from both sides and then dividing by x x (we can divide since x = 0 x=0 is not the root of the equation) we get a = b a=b , so there are no distinct a a and b b satysfying the above conditions.

Because a a and b b are the roots of x 2 + a x + 10 = 0 x^2+ax+10=0 and x 2 + b x + 10 = 0 x^2+bx+10=0 ,

a 2 + a b + 10 = 0 a^2+ab+10=0

b 2 + a b + 10 = 0 b^2+ab+10=0

Adding the equations we have ( a + b ) 2 = 20 (a+b)^2=-20 . Technically, there are complex values of a a and b b : a + b = ± 2 i 5 a+b= \pm 2i\sqrt5 , but there is no such answer among the choices.

Corrrect!!!!

Pi Han Goh - 4 years, 5 months ago

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