How can you solve 9x^4-24^3+16x^2=0?
I will assume your middle term is 24x^3
9x^4 - 24x^3 + 16x^2 = 0
x^2( 9x^2 - 24x + 16) = 0
mmmhh, the first and last terms are perfect squares, could it be ... ???
sure enough:
x^2(3x - 4)^2 = 0
x = 0 , x = ±2/√3
I knw it. . . X ^2 (9x^2 -24x +4 ) =0. . X^2(3x-4)(3x -4) =0. . 3x -4 =0/x^2. 3x-4 =0. . X =4/3 twice
To solve the equation 9x^4 - 24x^3 + 16x^2 = 0, you can use factoring.
Step 1: Factor out the greatest common factor (GCF) from each term. In this case, the GCF is x^2, which gives us:
x^2(9x^2 - 24x + 16) = 0
Step 2: Now, we can solve each factor separately.
First Factor: x^2 = 0
To solve this factor, set x^2 equal to zero and solve for x:
x^2 = 0
x = 0
Second Factor: 9x^2 - 24x + 16 = 0
To solve this factor, you can use the quadratic formula or try factoring it further. In this case, let's try factoring:
We need two numbers that multiply to give 16 and add up to -24. Those numbers are -4 and -4 (since -4 * -4 = 16 and -4 + -4 = -8). So we can rewrite the equation as:
9x^2 - 4x - 4x + 16 = 0
Now, we can factor by grouping:
(9x^2 - 4x) + (-4x + 16) = 0
x(9x - 4) - 4(9x - 4) = 0
Now, factor out the common binomial (9x - 4):
(9x - 4)(x - 4) = 0
Setting each factor equal to zero, we have two additional solutions:
9x - 4 = 0
9x = 4
x = 4/9
x - 4 = 0
x = 4
Therefore, the solutions to the equation 9x^4 - 24x^3 + 16x^2 = 0 are:
x = 0, x = 4/9, and x = 4.