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Solving equations with degree higher than 2

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Solving equations with degree higher than 2  [#permalink]

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New post 08 Feb 2019, 22:09
Please tell me what is wrong with the following procedure:

Q. Find solutions of the following expression.

(¼)x^4 - (¾)x^2 - 1 = 0

=> (¼)x^4 - (¾)x^2 = 1 -----------------moving 1 to RHS

=> (¼)*(x^2) * (x^2 - 3) = 1 ---------- taking (¼)*(x^2) common

=> (x^2)*(x^2 -3) = 4 ---------------- mulitplying 4 on both sides



=> x^2 = 4 OR x^2 - 3 = 4

=> x= +/-2 OR x = +/-7



Thus, there are four solutions: +2, -2, +7, -7.

However, the correct answer is that there are only two solutions: +2,-2.
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Re: Solving equations with degree higher than 2  [#permalink]

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New post 08 Feb 2019, 22:40
jk11 wrote:
Please tell me what is wrong with the following procedure:

Q. Find solutions of the following expression.

(¼)x^4 - (¾)x^2 - 1 = 0

=> (¼)x^4 - (¾)x^2 = 1 -----------------moving 1 to RHS

=> (¼)*(x^2) * (x^2 - 3) = 1 ---------- taking (¼)*(x^2) common

=> (x^2)*(x^2 -3) = 4 ---------------- mulitplying 4 on both sides



=> x^2 = 4 OR x^2 - 3 = 4

=> x= +/-2 OR x = +/-7



Thus, there are four solutions: +2, -2, +7, -7.

However, the correct answer is that there are only two solutions: +2,-2.


Hey jk11

you could have either substituted x^2 as some variable lets say a, or you could have just used it as it is, said that, i took the latter

x^2 will be +ive, x^4 will be positive

x^4 - 3x^2 - 4 = 0

x^2 (x^2 -4) + 1 (x^2-4) = 0

two solutions x^2 = -1 or x^2 = 4
only this is possible

x^2 = 4

two solutions -2 or 2
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Re: Solving equations with degree higher than 2  [#permalink]

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New post 09 Feb 2019, 08:19
KanishkM wrote:
jk11 wrote:
Please tell me what is wrong with the following procedure:

Q. Find solutions of the following expression.

(¼)x^4 - (¾)x^2 - 1 = 0

=> (¼)x^4 - (¾)x^2 = 1 -----------------moving 1 to RHS

=> (¼)*(x^2) * (x^2 - 3) = 1 ---------- taking (¼)*(x^2) common

=> (x^2)*(x^2 -3) = 4 ---------------- mulitplying 4 on both sides



=> x^2 = 4 OR x^2 - 3 = 4

=> x= +/-2 OR x = +/-7



Thus, there are four solutions: +2, -2, +7, -7.

However, the correct answer is that there are only two solutions: +2,-2.


Hey jk11

you could have either substituted x^2 as some variable lets say a, or you could have just used it as it is, said that, i took the latter

x^2 will be +ive, x^4 will be positive

x^4 - 3x^2 - 4 = 0

x^2 (x^2 -4) + 1 (x^2-4) = 0

two solutions x^2 = -1 or x^2 = 4
only this is possible

x^2 = 4

two solutions -2 or 2



I understand and agree with your approach. I just wanted to know what is wrong with mine? Which step made my solution wrong?
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Re: Solving equations with degree higher than 2  [#permalink]

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New post 09 Feb 2019, 08:24
jk11 wrote:
Please tell me what is wrong with the following procedure:

Q. Find solutions of the following expression.

(¼)x^4 - (¾)x^2 - 1 = 0

=> (¼)x^4 - (¾)x^2 = 1 -----------------moving 1 to RHS

=> (¼)*(x^2) * (x^2 - 3) = 1 ---------- taking (¼)*(x^2) common

=> (x^2)*(x^2 -3) = 4 ---------------- mulitplying 4 on both sides



=> x^2 = 4 OR x^2 - 3 = 4

=> x= +/-2 OR x = +/-7



Thus, there are four solutions: +2, -2, +7, -7.

However, the correct answer is that there are only two solutions: +2,-2.


x^2 = 7

will be x = \(\sqrt{7}\)

So the question would be asking for integer values

There was nothing wrong in the approach, I prefer not working with fractions, more scope of error :)
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If you notice any discrepancy in my reasoning, please let me know. Lets improve together.

Quote which i can relate to.
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Re: Solving equations with degree higher than 2  [#permalink]

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New post 10 Feb 2019, 13:09
The problem arises after the step where you arrive

x^2((x^2)-3) = 4

After this, you take x^2 = 4 or (x^2)-3 = 4
I don't know how you arrived at this step but let us go with it for once.

If x^2 = 4 , then (x^2)-3 = 4-3 = 1
So therefore the product of both of the terms is 4*1=4.

Now if (x^2)-3 = 4, then x^2 = 4+3 = 7 (using the same equation)
So therefore the product of the terms is 7*4=28.
Hence wrong
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Re: Solving equations with degree higher than 2   [#permalink] 10 Feb 2019, 13:09
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