3x^4-31x^2+28=0 Solve the equation??
To solve the equation 3x^4 - 31x^2 + 28 = 0, we can use a technique called factoring by substitution.
Step 1: Substitute a variable to simplify the equation. Let's substitute x^2 with a new variable, let's say y. Therefore, the equation becomes 3y^2 - 31y + 28 = 0.
Step 2: Factor the quadratic equation. The equation factors to (y - 4)(3y - 7) = 0.
Step 3: Set each factor equal to zero and solve for y.
Setting y - 4 = 0, we get y = 4.
Setting 3y - 7 = 0, we get y = 7/3.
Step 4: Substitute the original variable back in. Since we substituted x^2 with y, we can now substitute y back with x^2 in the solutions we found.
When y = 4, we have x^2 = 4. Taking the square root of both sides, we get x = ±2.
When y = 7/3, we have x^2 = 7/3. Taking the square root of both sides, we get x = ±√(7/3).
Therefore, the solutions to the equation 3x^4 - 31x^2 + 28 = 0 are x = ±2 and x = ±√(7/3).
you will need to use the quadratic equation
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let y = x^2
then
3 y^2 -31 y + 28 = 0
(3y-28)((y-1) = 0
y = 1 or 28/3
so
x^2 = 1
x = 1 or -1
x^2 = 28/3
x = 2 sqrt(7/3) or -2sqrt(7/3)
let x^2 = a
so we have
3a^2 - 31a + 28 = 0
(3a-28)(a-1) = 0
a = 28/3 or a = 1
so
x^2 = 28/3
x = ± 2√7/√3 or a = ±1
or
x = ± 2√21/3 , a = ±1