I am trying to solve the following using the Quadratic Formula but I am confused. Can you explain the steps?
Solve (find all complex-number solutions):
t^2 + 3 = 6t
Thank you
first of all, write it in its general form
ax^2 + bx + c = 0
t^2 - 6t + 3 = 0
a = 1, b = -6 , c = 3
t = (-b ±√(b^2-4ac) )/(2a)
= (6 ± √(36-4(1)(3))/2
= (6 ± √24)/2
= (6 ± 2√6)/2
= 3 ± √6
No complex roots here.
Reiny,
I see it now, you took the =6t and made it a -6t for b. That was what was confusing me. Thank you so much, now I understand
Sure! To solve the given equation using the Quadratic Formula, you need to follow a series of steps. Here's the breakdown:
Step 1: Rewrite the equation in standard quadratic form (ax^2 + bx + c = 0). In this case, the equation is already in standard form.
Step 2: Identify the coefficients a, b, and c. In our equation, a = 1, b = -6, and c = -3.
Step 3: Plug the values of a, b, and c into the Quadratic Formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Step 4: Substitute the values into the formula:
t = (-(6) ± √((6)^2 - 4(1)(-3))) / (2(1))
Step 5: Simplify the equation:
t = (-6 ± √(36 + 12)) / 2
t = (-6 ± √(48)) / 2
t = (-6 ± 4√3) / 2
Step 6: Simplify further:
t = -3 ± 2√3
So the complex-number solutions to the equation t^2 + 3 = 6t are t = -3 + 2√3 and t = -3 - 2√3.
I hope this helps! Let me know if you have any further questions.