How do you find the roots, real and imaginary, of # h(t)=-2t^2+16t -4 # using the quadratic formula?

Answer 1

#a=-2,b=16,c=-4 ->x=(-b+- (sqrt(b^2-4ac)))/(2a) = (-16+- sqrt(16^2-4(-2)(-4)))/(2(-2)) = (-16+-sqrt224)/-4 = (-16+-4sqrt14)/-4 = 4+- -sqrt14#

List a,b,c and then substitute in to the quadratic formula then solve for x

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Kennedy Adams

To find the roots of the quadratic equation ( h(t) = -2t^2 + 16t - 4 ) using the quadratic formula:

Identify the coefficients: ( a = -2 ), ( b = 16 ), and ( c = -4 ).
Substitute the values of ( a ), ( b ), and ( c ) into the quadratic formula: ( t = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} )
Plug in the values: ( t = \frac{{-16 \pm \sqrt{{16^2 - 4(-2)(-4)}}}}{{2(-2)}} )
Simplify inside the square root: ( t = \frac{{-16 \pm \sqrt{{256 - 32}}}}{{-4}} ) ( t = \frac{{-16 \pm \sqrt{{224}}}}{{-4}} )
Further simplify if possible: ( t = \frac{{-16 \pm 4\sqrt{{14}}}}{{-4}} ) ( t = 4 \pm \sqrt{{14}} )

Therefore, the roots of the equation are ( t = 4 + \sqrt{{14}} ) and ( t = 4 - \sqrt{{14}} ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.