(A)Write the equation in standard form and calculate its discriminant. (B)Solve the equation by using the quadratic formula. (C)After solving the equation, write it in factored form. 0=8+4t-t^2 Please help!!!!! :(
so, what is the standard form? If you can't get that, you're hosed.
Maybe I have to switch sides so it might be t^2-4t+8=0. Please help me more!!!! :(
Unless you make an error in the "switching"
0=8+4t-t^2 ----> t^2 - 4t - 8 = 0
hint: use the formula, or complete the square
Sure! Let's go step by step to solve the equation.
(A) To write the equation in standard form, we need to arrange it in descending order of the variable and set it equal to zero.
The given equation is: 0 = 8 + 4t - t^2.
Rearranging the terms, we get: 0 = -t^2 + 4t + 8.
So, the equation in standard form is -t^2 + 4t + 8 = 0.
Now, let's calculate the discriminant.
The discriminant (D) of a quadratic equation in the form ax^2 + bx + c = 0 is given by the formula D = b^2 - 4ac.
Comparing with our equation, we have a = -1, b = 4, and c = 8.
Substituting these values into the formula, we get D = (4)^2 - 4(-1)(8).
Calculating further, we have D = 16 + 32 = 48.
Therefore, the discriminant is 48.
(B) To solve the equation using the quadratic formula, we can use the formula x = (-b ± √(b^2 - 4ac)) / (2a).
In our equation, a = -1, b = 4, and c = 8.
Substituting these values into the formula, we get:
t = (-4 ± √(4^2 - 4(-1)(8))) / (2(-1)).
Simplifying further, we have:
t = (-4 ± √(16 + 32)) / (-2).
t = (-4 ± √48) / (-2).
Now, simplifying inside the square root, we get:
t = (-4 ± √(16 * 3)) / (-2).
t = (-4 ± 4√3) / (-2).
Simplifying the expression, we have:
t = 2 ± 2√3.
Therefore, the solutions to the equation are t = 2 + 2√3 and t = 2 - 2√3.
(C) Finally, let's write the equation in factored form.
From the solutions obtained in step (B), we can write the equation as:
(t - 2 - 2√3)(t - 2 + 2√3) = 0.
This is the factored form of the equation.
I hope this explanation helps you to solve the equation. If you have any further questions, feel free to ask!