Answer:
=-1
=6
Step-by-step explanation:
By factoring:
T²-5t-6=0
(t - 6) (t + 1)
t - 6 = 0. t + 1= 0
t= 6 t= -1
It tok me lot f time to finish solving this haha
Factoring t2-5t+6
The first term is, t2 its coefficient is 1 .
The middle term is, -5t its coefficient is -5 .
The last term, "the constant", is +6
Step-1 : Multiply the coefficient of the first term by the constant 1 • 6 = 6
Step-2 : Find two factors of 6 whose sum equals the coefficient of the middle term, which is -56
-6 + -1 = -7
-3 + -2 = -5
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -3 and -2
t2 - 3t - 2t - 6
Step-4 : Add up the first 2 terms, pulling out like factors :
t • (t-3)
Add up the last 2 terms, pulling out common factors :
2 • (t-3)
Step-5 : Add up the four terms of step 4 :
(t-2) • (t-3)
Which is the desired factorization
equation at the end of step 1;
(t - 2) • (t - 3) = 0
Step 2 Theory roots of a product;
A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Solve Quadratic equation using quadratic formula;
olving t2-5t+6 = 0 by the Quadratic Formula .
According to the Quadratic Formula, t , the solution for At2+Bt+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
t = ————————
2A
In our case, A = 1
B = -5
C = 6
Accordingly, B2 - 4AC =
25 - 24 =
1
Applying the quadratic formula :
5 ± √ 1
t = ————
2
So now we are looking at:
t = ( 5 ± 1) / 2
Two real solutions:
t =(5+√1)/2= 3.000
or:
t =(5-√1)/2= 2.000
the two solutions;
1. t=3
2. t=2
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Answers & Comments
Answer:
=-1
=6
Step-by-step explanation:
By factoring:
T²-5t-6=0
(t - 6) (t + 1)
t - 6 = 0. t + 1= 0
t= 6 t= -1
Answer:
It tok me lot f time to finish solving this haha
Step-by-step explanation:
Factoring t2-5t+6
The first term is, t2 its coefficient is 1 .
The middle term is, -5t its coefficient is -5 .
The last term, "the constant", is +6
Step-1 : Multiply the coefficient of the first term by the constant 1 • 6 = 6
Step-2 : Find two factors of 6 whose sum equals the coefficient of the middle term, which is -56
-6 + -1 = -7
-3 + -2 = -5
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -3 and -2
t2 - 3t - 2t - 6
Step-4 : Add up the first 2 terms, pulling out like factors :
t • (t-3)
Add up the last 2 terms, pulling out common factors :
2 • (t-3)
Step-5 : Add up the four terms of step 4 :
(t-2) • (t-3)
Which is the desired factorization
equation at the end of step 1;
(t - 2) • (t - 3) = 0
Step 2 Theory roots of a product;
A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Solve Quadratic equation using quadratic formula;
olving t2-5t+6 = 0 by the Quadratic Formula .
According to the Quadratic Formula, t , the solution for At2+Bt+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
t = ————————
2A
In our case, A = 1
B = -5
C = 6
Accordingly, B2 - 4AC =
25 - 24 =
1
Applying the quadratic formula :
5 ± √ 1
t = ————
2
So now we are looking at:
t = ( 5 ± 1) / 2
Two real solutions:
t =(5+√1)/2= 3.000
or:
t =(5-√1)/2= 2.000
the two solutions;
1. t=3
2. t=2