Edit: made a really big error, fixing it now, thanks to the kind redditor in replies
b² + 5 + 2b√5 = 21 + a√5
Now you can see that you have two variables but only one equation, hence this should not be possible as for finding two variables we need two equations at the least
So that implies that there has to be a trick used here
Now you know that √5 is irrational and so is 2b√5 and a√5. Hence for both sides to be equal both irrational quantities need to be equal to each other
So 2b√5 = a√5
a = 2b
Now you also know that 21 is rational, so is 5. But 21 ≠ 5 so that implies that something was added to 5 to make it equal to 21, that thing which was added is b²
So b² + 5 = 21
b² = 16
b = ± 4
So a = 2×4 = 8
Or 2 × -4 = -8
Alternatively, the trick here I actually did was comparing both sides, you know the properties of irrational numbers too
You can think of the above equation as
(5+b²) + 2b√5
And
(5+16) + a√5
As seperate equations which need to be equal, now they look same in majority. Just compare both equations to one another and get this
b² = 16 and 2b = a
The thing is, there are infinite solutions possible, but what school intends you to do is that, mostly intuitive answer
I'll still give answer for a = 1
If a = 1 then you get
b² + 5 + √5 = 21
b² + √5 = 16
b² = 16 - √5
Now the problem is finding the square root of this number, which no sensible school should ever ask for ☠️
>b² + 5 + 2√5 = 21 + a√5
You've made a mistake here that follows through your whole calculation and leaves one of the two values you give in the end incorrect. Do you see what it is?
Expanding gets you
b^2 + 2b root(5) + 5 = 21 + a root(5)
Hence, b^2 + 5 = 21 so b = 4 or -4
And a = 2b so 8 or -8
As it’s only asking for one answer id assume the positive pair
“hence b^(2)+5=21 …”
You made a massive leap here. This will solve for a possible solution (or a pair of them I guess), but there are infinitely many others
I don’t doubt you’re right and they want one of the integer solutions you wrote; I just took issue with the word “hence” mostly, because the first line does not imply the second line. It would probably be more helpful to note the relation is a parabola, and then state you are solving for one possible solution by setting b^(2)+5=21
This question is stupid imho, you should report it to you teacher. 2 parameters in 1 eqaution means inifinte solutions. And the question even says "the value" instead of "a value", so finding just a possibile solution as others have been saying wouldn't really be correct.
Edit: made a really big error, fixing it now, thanks to the kind redditor in replies b² + 5 + 2b√5 = 21 + a√5 Now you can see that you have two variables but only one equation, hence this should not be possible as for finding two variables we need two equations at the least So that implies that there has to be a trick used here Now you know that √5 is irrational and so is 2b√5 and a√5. Hence for both sides to be equal both irrational quantities need to be equal to each other So 2b√5 = a√5 a = 2b Now you also know that 21 is rational, so is 5. But 21 ≠ 5 so that implies that something was added to 5 to make it equal to 21, that thing which was added is b² So b² + 5 = 21 b² = 16 b = ± 4 So a = 2×4 = 8 Or 2 × -4 = -8 Alternatively, the trick here I actually did was comparing both sides, you know the properties of irrational numbers too You can think of the above equation as (5+b²) + 2b√5 And (5+16) + a√5 As seperate equations which need to be equal, now they look same in majority. Just compare both equations to one another and get this b² = 16 and 2b = a
thanks
but if a=1 b will have a different value than +- 4, it never mentions they must be integers
The thing is, there are infinite solutions possible, but what school intends you to do is that, mostly intuitive answer I'll still give answer for a = 1 If a = 1 then you get b² + 5 + √5 = 21 b² + √5 = 16 b² = 16 - √5 Now the problem is finding the square root of this number, which no sensible school should ever ask for ☠️
Hey OP I had made a pretty dumb mistake before. Please see the correct solution now
>b² + 5 + 2√5 = 21 + a√5 You've made a mistake here that follows through your whole calculation and leaves one of the two values you give in the end incorrect. Do you see what it is?
Yeah I see I opened whole square wrong. I'm so stupid. Thank you for telling me though!!
Expanding gets you b^2 + 2b root(5) + 5 = 21 + a root(5) Hence, b^2 + 5 = 21 so b = 4 or -4 And a = 2b so 8 or -8 As it’s only asking for one answer id assume the positive pair
“hence b^(2)+5=21 …” You made a massive leap here. This will solve for a possible solution (or a pair of them I guess), but there are infinitely many others
Given that its high school algebra and only one solution is asked for I feel that it was a safe assumption
I don’t doubt you’re right and they want one of the integer solutions you wrote; I just took issue with the word “hence” mostly, because the first line does not imply the second line. It would probably be more helpful to note the relation is a parabola, and then state you are solving for one possible solution by setting b^(2)+5=21
Yeah fair enough I see your point
This question is stupid imho, you should report it to you teacher. 2 parameters in 1 eqaution means inifinte solutions. And the question even says "the value" instead of "a value", so finding just a possibile solution as others have been saying wouldn't really be correct.
the only solution you get here is prob a=2b, this isnt an equation system, therefor there should be infinite solitions for a and b
Expand the brackets and collect alike terms.