Solving quadratic eqn: (1/2)t^2 - 5t + 25 = -t^2 + 6t + 16

[wintertwothousand]

Here is the problem:

(1/2)t^2 - 5t + 25 = -t^2 + 6t + 16

I know I must subtract the right side, thereby setting it to 0, but when I do this I get 1.5t^2 - 11t + 9 = 0. How is it possible to plug this into a quadratic formula when I have a negative for my square root?

The answer key says t= 0.94 how did they find this??

 

[skeeter]

Re: Solving a quadratic equation

\(\displaystyle 1.5t^2 - 11t + 9 = 0\)

\(\displaystyle b^2 - 4ac = (-11)^2 - 4(1.5)(9) = 121 - 54 = 67\)

\(\displaystyle t = \frac{11 \pm \sqrt{67}}{3}\)

t = .94 is one solution , there is another.

 

[Mrspi]

Re: Solving a quadratic equation

Here is the problem:

(1/2)t^2 - 5t + 25 = -t^2 + 6t + 16

I know I must subtract the right side, thereby setting it to 0, but when I do this I get 1.5t^2 - 11t + 9 = 0. How is it possible to plug this into a quadratic formula when I have a negative for my square root?

The answer key says t= 0.94 how did they find this??

Your equation,

1.5t^2 - 11t + 9 = 0

is correct.

This is a quadratic equation of the form

ax^2 + bx + c = 0,
where a = 1.5, b = -11, and c = 9.

Substitute into the quadratic formula, and do the arithmetic.

I'm sorry...I don't understand this question: How is it possible to plug this into a quadratic formula when I have a negative for my square root?

If you're talking about the part of the quadratic formula which says sqrt(b^2 - 4ac), then you'll have sqrt((-11)^2 - 4*1.5*9), or sqrt(121 - 54)...that's not going to be negative.

 

[wintertwothousand]

Re: Solving a quadratic equation

Ah! I was missing a parenthesis around the 11, which explains why I was getting -121.

Thanks for the help.

 

[mmm4444bot]

Ah! I was missing [parentheses] around the 11, …

Were you using a calculator, instead of your brain? (My brain does not require parentheses.)