Question 430115: t^2+4t-2=0
Answer by haileytucki(390)   (Show Source):
You can put this solution on YOUR website! NOTE: ~ SYMBOLIZES THE SQUARE ROOT OF
t^(2)+4t-2=0
Use the quadratic formula to find the solutions. In this case, the values are a=1, b=4, and c=-2.
t=(-b\~(b^(2)-4ac))/(2a) where at^(2)+bt+c=0
Use the standard form of the equation to find a, b, and c for this quadratic.
a=1, b=4, and c=-2
Substitute in the values of a=1, b=4, and c=-2.
t=(-4\~((4)^(2)-4(1)(-2)))/(2(1))
Simplify the section inside the radical.
t=(-4\2~(6))/(2(1))
Simplify the denominator of the quadratic formula.
t=(-4\2~(6))/(2)
First, solve the + portion of \.
t=(-4+2~(6))/(2)
Factor out the GCF of 2 from each term in the polynomial.
t=(2(-2)+2(~(6)))/(2)
Factor out the GCF of 2 from -4+2~(6).
t=(2(-2+~(6)))/(2)
Reduce the expression (2(-2+~(6)))/(2) by removing a factor of 2 from the numerator and denominator.
t=(-2+~(6))
Remove the parentheses around the expression -2+~(6).
t=-2+~(6)
Next, solve the - portion of \.
t=(-4-2~(6))/(2)
Factor out the GCF of 2 from each term in the polynomial.
t=(2(-2)+2(-~(6)))/(2)
Factor out the GCF of 2 from -4-2~(6).
t=(2(-2-~(6)))/(2)
Reduce the expression (2(-2-~(6)))/(2) by removing a factor of 2 from the numerator and denominator.
t=(-2-~(6))
Remove the parentheses around the expression -2-~(6).
t=-2-~(6)
The final answer is the combination of both solutions.
t=-2+~(6),-2-~(6)_t= 0.4494897,-4.44949
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