How do you solve t to the 2nd power - 10t + 21

Question

Can u answer ASAP I really need the answer for this 

Answer 1

Hi Daniela;

t² - 10t + 21

You can also write this as...

t^2-10t+21

For the factor, we need to FOIL.

For the FOIL...

FIRST must be (t)(t)=t²

OUTER and INNER must add-up to 10t.

LAST must be (7)(3) or (3)(7) or (21)(1) or (1)(21) and both numbers must be negative to render OUTER and INNER sum as negative, while the product of LAST as positive.

(t-7)(t-3)

Let's FOIL...

FIRST...(t)(t)=t²

OUTER...(t)(-3)=-3t

INNER...(-7)(t)=-7t

LAST...(-7)(-3)=21

t²-3t-7t+21

t²-10t+21

(t-7)(t-3)

To solve, we must set this equal to zero...

(t-7)(t-3)=0

Either or both parenthetical equations must equal zero...

t-7=0

t=7

t-3=0

t=3

 

 

Answer 2

t^2 - 10t + 21

 

So you need to FACTOR this quadratic ... the easiest way is to realize that since the FINAL value (+21) is

positive, it either has 2 negative or 2 positive factors. NOW look at the MIDDLE term, since it is NEGATIVE - this tells you that you want to make the 2 factors NEGATIVE.

 

SO we have (t - ?)(t - ?) ... so what are 2 factors of 21 that ADD UP TO 10?? 

 

hopefully you get 3 and 7

 

That gives us the factors

(t - 3)(t - 7)

 

We 'solve' the problem by setting these each to ZERO and finding t

 

t - 3 = 0      t - 7 = 0

  +3    +3      +7    +7

t = 3           t = 7

 

The SOLUTIONS for t^2 - 10t + 21 = 0 are t=3 and t=7