Find all real solutions of the equation by first rewriting the equation as a quadratic equation. (Enter your answers as a comma-separated list.)
4x4 + 15x2 − 4 = 0
Let y=x²
the equation can then be written as
4y²+15y-4=0
Solve for y using the quadratic formula (i.e. find y1 and y2).
Since y=x², then
x=±√(y1) or ±√(y2)
giving a total of 4 solutions.
Find all real solutions of the equation by first rewriting the equation as a quadratic equation. (Enter your answers as a comma-separated list.)
4^4 + 15^2 − 4 = 0
To rewrite the equation as a quadratic equation, let's substitute a variable.
Let's substitute y = x².
The equation becomes:
4y² + 15y - 4 = 0.
Now we have a quadratic equation. To solve it, we can use the quadratic formula:
y = (-b ± √(b² - 4ac)) / (2a)
In this case, a = 4, b = 15, and c = -4.
Plugging these values into the quadratic formula, we get:
y = (-15 ± √(15² - 4(4)(-4))) / (2(4))
Simplifying further, we have:
y = (-15 ± √(225 + 64)) / 8
y = (-15 ± √289) / 8
Now we can find the two possible values of y:
y₁ = (-15 + √289) / 8 = (8)/8 = 1
y₂ = (-15 - √289) / 8 = (-23)/8 = -2.875
Since y = x², we can substitute these values back into the equation:
For y = 1: x² = 1. Taking the square root of both sides, we get x = ±1.
For y = -2.875: x² = -2.875. Since we cannot take the square root of a negative number and obtain a real solution, this does not have any real solutions.
In conclusion, the real solutions of the equation 4x⁴ + 15x² - 4 = 0 are x = ±1.
To rewrite the equation as a quadratic equation, let's introduce a substitution. Let u = x^2. Then, we have:
4u^2 + 15u − 4 = 0
Now, we can solve this quadratic equation for u using the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a = 4, b = 15, and c = -4 into the quadratic formula:
u = (-(15) ± √((15)^2 - 4(4)(-4))) / (2(4))
Simplifying:
u = (-15 ± √(225 + 64)) / 8
u = (-15 ± √289) / 8
u = (-15 ± 17) / 8
So, we have two possible values for u:
u1 = (-15 + 17) / 8 = 2/8 = 1/4
u2 = (-15 - 17) / 8 = -32/8 = -4
Now, we can substitute back x^2 for u:
For u1:
x^2 = 1/4
x = ± √(1/4)
x = ± 1/2
For u2:
x^2 = -4
This equation does not have real solutions because the square of a real number cannot be negative.
Therefore, the real solutions of the equation 4x^4 + 15x^2 - 4 = 0 are x = 1/2 and x = -1/2.