Find all real and/or imaginary solutions.
x4+5x2−14=0
(1 point)
Responses
x=±2–√i or x=±7–√
x=±2–√i or x=±7–√
x=±3–√i or x=±6–√
x=±3–√i or x=±6–√
x=±7–√i or x=±2–√
x=±7–√i or x=±2–√
x=±6–√i or x=±3–√
To find the solutions to the equation, we can use the quadratic formula. Let's substitute y = x^2:
y^2 + 5y - 14 = 0
Using the quadratic formula, we have:
y = (-b ± √(b^2 - 4ac))/(2a)
a = 1, b = 5, c = -14
y = (-5 ± √(5^2 - 4(1)(-14))) / (2(1))
y = (-5 ± √(25 + 56)) / 2
y = (-5 ± √81) / 2
y = (-5 ± 9) / 2
This gives us two possible values for y:
y1 = (-5 + 9) / 2 = 2
y2 = (-5 - 9) / 2 = -7
Now we can back substitute to find the solutions for x:
y1 = x^2 = 2
x^2 - 2 = 0
(x - √2)(x + √2) = 0
x = ±√2
y2 = x^2 = -7
x^2 + 7 = 0
(x + √7i)(x - √7i) = 0
x = ±√7i
So the solutions are x = ±√2 and x = ±√7i.