# Find the roots of x^4 – 13x^2 + 36 = 0

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to find the roots of x^4 – 13x^2 + 36 = 0. As we have x raised to the power 4, we are going to have 4 roots.

Now let us see the given equation :

x^4 – 13x^2 + 36 = 0

we see that -13 can be written as -4 – 9 so that the sum is -13 and the product is 36

=> x^4 – 4x^2 – 9x^2 + 36 = 0

=> x^2 ( x^2 – 4) – 9( x^2 – 4) = 0

=> (x^2 – 4)(x^2 – 9) =0

Now x^2 – 4 and x^2 – 9 can be factorized using the relation x^2 – y^2 = (x – y)*(x + y)

=> (x – 2)(x + 2)(x – 3)(x + 3) =0

So for (x – 2) = 0, we have x = 2

for (x + 2) = 0, we have x = -2

for (x – 3) = 0 we have x = 3

and for (x + 3) =0, we have x = -3

Therefore the roots of x^4 – 13x^2 + 36 = 0 are x = 2, x = -2, x = 3 and x = -3.

changchengliang | Elementary School Teacher | (Level 2) Adjunct Educator

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The task is to:

Find the roots of x^4 – 13x^2 + 36 = 0

We can make the equation look friendlier by letting x^2=y.

Then the equation becomes:

y^2 - 13y + 36 = 0

Keeping in mind to obtain the number 13 from addition of factors of 36, and knowing that 36 can be obtained by multiplying 9 and 4, the factorization becomes:

(y-9)(y-4) = 0

Hence y=9 or 4.

That translate to x^2=9 => x=+3 or -3

or    x^2=4 =>  x=+2 or -2

Therefore, the roots of the equation are:

x1 = -3

x2 = -2

x3 = 2

x4 = 3

To appreciate the solution graphically, click on the link at the bottom of this answer

pilgrimpilgrim | Student, Undergraduate | eNotes Newbie

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

Using a chain of equivalent auxiliary problems and transforming the original condition into another condition:

(2x^2)^2 -2(2x^2)13+144=0

Working in the same direction and transforming yet again into another condition:

(2x^2)-2(2x^2)13+169=25

Proceeding in the same way, we obtain

(2x^2-13)^2=25

2x^2-13=+-5

x^2=13+-5/2

x=+-sqrt 13+-5/2

x=3, or -3, or 2, or -2

:Courtest of "How to Solve It" - George Polya.

neela | High School Teacher | (Level 3) Valedictorian

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To  find the roots x^4-13x^2+36 = 0.

We put x^2= t. Then x^4 = t^2. With this transformation, the given expression x^4-13x^2+26 becomes t^2-13t+36 , a quadratic polynomial in t.

The middle term -13t = -9t - 4t and  (-9t)(-4t) = product of first and 3rd terms = x^2*36 = 3x^2. Therefore we write

t^2 -13t +36 = 0.

t^2-9t -4t +36 0.

t^2 -13t +36 = 0.

t(t-9) -4(t-9) = 0.

t^2 -13t +36 = 0.

(t-9)(t-4) = 0

Nowe we replace t by x^2 and get:

t^2 -13t +36 = x^4 – 13x^2 + 36 = (x^2-9)(x^2-4).

x^4 – 13x^2 + 36 = (x^2-9)(x^2-4) = 0.

x^2 -9 = 0 gives x^2 = 9, x = 3 or x = -3.

x^2-4 = 0 gives x^2 = 4,. So x= 2 or x = -2.

Therefore  solution for x are :  - 3 , -2 , 2, 3.

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