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Solve `x^4 -13x^2 +36=0` `Thanks` ` `

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phanpal999 | Student, Grade 8 | (Level 1) Salutatorian

Posted June 21, 2013 at 12:21 AM via web

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Solve `x^4 -13x^2 +36=0`

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llltkl | College Teacher | (Level 3) Valedictorian

Posted June 21, 2013 at 12:45 AM (Answer #1)

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Given `x^4-13x^2+36=0`

To solve for x, first factorize by splitting the middle term.  -13 can be written as the sum of 2 numbers which give a product 36. This can be done as -13 = -9 +(-4). So,

`x^4-9x^2-4x^2+36=0`

`rArr x^2(x^2-9)-4(x^2-9)=0`

`rArr (x^2-9)(x^2-4)=0`

`rArr {(x)^2-(3)^2}{(x)^2-(2)^2}=0`

Applying `a^2-b^2=(a+b)(a-b)` we get:

`(x+3)(x-3)(x+2)(x-2)=0`

Therefore, x=-3,+3,-2,+2.

Sources:

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zach2794 | Student, Undergraduate | (Level 1) eNoter

Posted June 22, 2013 at 3:07 AM (Answer #2)

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`x^4-13x^2+36=0`` `

We can split this up into:

`(x^2-9)*(x^2-4)=0`

Then with this being true we also know that

`(x^2-9)=0` and `(x^2-4)=0`

so solving for x we get that `x^2=9,4`

Taking the square root of both sides gives us:

Answer:`` `x=+-3,+-2`

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