find the zeros of the function algebraically f(x)= 49x^4- 36x^2

Question

i just dont understand, ive tried and tried graphing it in my calculator and tried to find the zeros but i cant figure it out

Answer 1

 Just Simply Use Identity:

  

   a^2 - b^2 = ( a+b) ( a-b)

 

    Then:

 

      ( 49X^4 - 36 X^2) = 0

 

        (7 X^2 + 6 X) ( 7X^2 - 6X) =0

 

        7X^2 + 6X = 0, then           X( 7X +6) =0    X=0,  7x +6 =0   X = -6/7

 

            7X^2 -6x =0                   X( 7x - 6) =0   X =0   7x -6 =0    X =6/7

Answer 2

Hi Elizabeth;

f(x)= 49x⁴- 36x²

For the FOIL...

FIRST must be (7x²)(7x²)=49x⁴.

OUTER and INNER must add-up to zero therefore FIRST and LAST must be identical except for the fact that one item of LAST must be negative.

LAST must be (6x)(6x)=36x².

0=(7x²+6x)(7x²-6x)

Let's FOIL...

FIRST...(7x²)(7x²)=49x⁴

OUTER...(7x²)(-6x)=-42x³

INNER...(6x)(7x²)=42x³

LAST...(6x)(-6x)=-36x²

0=49x⁴-42x³+42x³-36x²

0=49x⁴-36x²

0=(7x²+6x)(7x²-6x)

Either or both parenthetical equation(s) equal zero...

0=(7x²+6x)

0=x(7x+6)

Either x or (7x+6) must equal zero...

x=0

0=7x+6

-6=7x

-6/7=x

0=7x²-6x

0=x(7x-6)

x=0

0=7x-6

6=7x

6/7=x

 

Answer 3

f(x) = 49x^4 - 36x^2 :  Both terms share an x^2 term so factor out that term

x^2 (49x^2 - 36) : Now the inner quantity can easily be factored

x^2 (7x + 6)(7x - 6) : From here it is possible to solve for the zeros, or values of x which make f(x) = 0

x = 0 (double root), x = -6/7, and x = 6/7