Subtract both given equations a^2-b^2=b-a. i . e (a+b).(a-b)=-1(a-b) i .e a=-1-b . Now substitute this in second equation b^2=-1-b+13. i .e b^2 +b-12=0 i .e (b-4).(b+3)=0 i .e b=(4,-3) now a can be calculated
2 cases: a = b, that's possible, since, at this way, the system is composed by 2 identical equations; a ≠ b. case a = b: a = b => a² = a + 13 => a² - a - 13 = 0 => a = 1/2(1±sqrt(53)) case a ≠ b a² = b + 13 & b² = a + 13 from this system is true that 13 = a² - b = b² - a, so we have: a² - b = b² - a a² - b² = b - a = -(a - b) a² - b² = (a-b)(a+b), so by substitution: (a-b)(a+b) = -(a-b) a + b = -1 => a = -(b+1) by substituting a in b² = a + 13, we obtain: b² = -b + 12 => b² + b - 12 = 0, from that b ={-4,3} so if we substitute b solutions in a = -(b+1) there's these couples of solutions: (a,b) ={(3,-4),(-4,3)} In conclusion we have these solutions, including all cases: (a,b) ={(3,-4),(-4,3),(1/2(1±sqrt(53)),1/2(1±sqrt(53)))} Yeah, I added also the other cases to find all real solutions
There are two irrational solutions not shown in the video. The solutions are: (a,b) = (((1+sqrt(53))/2), ((1+sqrt(53))/2)) and (((1-sqrt(53))/2), ((1-sqrt(53))/2)). You can plug these solutions back into the original equations and find that they are indeed valid solutions.
Oxford doesn't do admissions like that. Apart from that, the question is way too easy to ever to be considered in case Oxford would do admissions like this.
More creative to let a=x, b =y. Sketch the two parabolas, verify two solutions (and symmetry). Not hard to realise 13+3 = 4^2 is it? Especially if as claimed, it’s for Oxford interviews. Most Oxford Maths and Physics candidates would do this in their head!
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Subtract both given equations a^2-b^2=b-a. i . e (a+b).(a-b)=-1(a-b) i .e a=-1-b . Now substitute this in second equation b^2=-1-b+13. i .e b^2 +b-12=0 i .e (b-4).(b+3)=0 i .e b=(4,-3) now a can be calculated
There are two more solutions. I left a comment about it.
Actually you’re right, I forgot he said a =/= b
2 cases: a = b, that's possible, since, at this way, the system is composed by 2 identical equations; a ≠ b.
case a = b:
a = b => a² = a + 13 => a² - a - 13 = 0 => a = 1/2(1±sqrt(53))
case a ≠ b
a² = b + 13 & b² = a + 13
from this system is true that 13 = a² - b = b² - a, so we have:
a² - b = b² - a
a² - b² = b - a = -(a - b)
a² - b² = (a-b)(a+b), so by substitution:
(a-b)(a+b) = -(a-b)
a + b = -1 => a = -(b+1)
by substituting a in b² = a + 13, we obtain:
b² = -b + 12 => b² + b - 12 = 0, from that b ={-4,3}
so if we substitute b solutions in a = -(b+1) there's these couples of solutions:
(a,b) ={(3,-4),(-4,3)}
In conclusion we have these solutions, including all cases:
(a,b) ={(3,-4),(-4,3),(1/2(1±sqrt(53)),1/2(1±sqrt(53)))}
Yeah, I added also the other cases to find all real solutions
I agree your solution. Because I got the result like you.
a^2-13=b, so (a^2-13)^2=a+13.
Expanding, a^4-26a^2+169=a+13
Rearrange, giving us a^4-26a^2-a=-156
Allowing us to get a directly.
Well if this is oxford then India's 10 grade(high school final year) is way harder
{13ab+13ab ➖ }=26ab^2 2^13ab^2 2^13^1ab^2 2^1^1ab^2 1ab^2 (ab ➖ 2ab+1). {13ba+13ba ➖}=26ba^2 2^13ba^2 2^13^1ba^2 2^1^1ba^2 1ba^2 (ba ➖ 2ba+1).
There are two irrational solutions not shown in the video. The solutions are: (a,b) = (((1+sqrt(53))/2), ((1+sqrt(53))/2)) and (((1-sqrt(53))/2), ((1-sqrt(53))/2)). You can plug these solutions back into the original equations and find that they are indeed valid solutions.
Oh wait sorry you said they aren’t equal, my bad
a = -3 b =-4
No, A = +3. If it were -3 then B squared would have to be equal to 10.
Oxford doesn't do admissions like that. Apart from that, the question is way too easy to ever to be considered in case Oxford would do admissions like this.
q=b^2-13 , (b-3)(b^3+3b^2-17b-52)=0 , b=3 , b^3+3b^2-17b-52=0 , (b+4)(b^2-b-13)=0 , b=-4 ,
b^2-b-13=0 , b=(1+/-V53)/2 ,
b= 3 , -4 , case 1 , b=3 , a=b^2-13 , a=9-13 , b= -4 , case 2 , b=-4 , a=16-13 , b=3 ,
solu , (a , b) , (-4 , 3) , (3 , -4) , test , a^2=b+13 , c1 , (-4)^2=3+13 , 16=16 ,
c2 , 3^2= -4+13 , 9=9 , OK ,
There’s two more solutions
Actually you are right I forgot he said a not equal to b
Hi i have another method....
What is it? Share with us!
(1): a² = b + 13
(2): b² = a + 13
(1) - (2)
a² - b² = (b + 13) - (a + 13)
a² - b² = b + 13 - a - 13
a² - b² = b - a → recall: a² - b² = (a + b).(a - b)
(a + b).(a - b) = b - a
(a + b).(a - b) - (b - a) = 0
(a + b).(a - b) + (a - b) = 0
(a - b).[(a + b) + 1] = 0 → where: a ≠ b
a + b + 1 = 0
a + b = - 1 ← equation (3)
(1) + (2)
a² + b² = (b + 13) + (a + 13)
a² + b² = a + b + 26 → recall (3): a + b = - 1
a² + b² = 25 ← equation (4)
From (3):
a + b = - 1
(a + b)² = 1
a² + b² + 2ab = 1 → recall (4): a² + b² = 25
25 + 2ab = 1
2ab = - 24 ← equation (5)
(a - b)² = a² + b² - 2ab → recall (4): a² + b² = 25
(a - b)² = 25 - 2ab → recall (5): 2ab = - 24
(a - b)² = 25 + 24
(a - b)² = 49
a - b = ± 7 ← equation (6)
First case:
a - b = 7 → recall (3): a + b = - 1
a + b = - 1
--------------------------the sum
2a = 6
→ a = 3 → recall: a + b = - 1
→ b = - 4
Second case:
a - b = - 7 → recall (3): a + b = - 1
a + b = - 1
--------------------------the sum
2a = - 8
→ a = - 4 → recall: a + b = - 1
→ b = 3
There are two more solutions; see my comment in here
You are right actually, I forgot that he said a not equal to b
This is part of the entrance exam into Oxford? Are they letting elementary school kids apply these days? Boy is the bar set low.
You are complicating things.
More creative to let a=x, b =y. Sketch the two parabolas, verify two solutions (and symmetry). Not hard to realise 13+3 = 4^2 is it? Especially if as claimed, it’s for Oxford interviews. Most Oxford Maths and Physics candidates would do this in their head!
Неправильное решение.От начала до конца.
I solved this in my head in less than 2 minutes.
A = -4
B = 3
You made this way too complicated.
not the only solutions, you can verify
The solutions are restricted. a is NOT equal to b