solve t^2-3t-18=0 by factoring, by completing the square, and by using the quadratic formula. Compare your results.
To solve the quadratic equation t^2 - 3t - 18 = 0, we will solve it by factoring, completing the square, and using the quadratic formula.
1. Factoring:
First, we will look for two numbers whose product is -18 and sum is -3. The numbers -6 and 3 satisfy these conditions.
So, t^2 - 3t - 18 = (t - 6)(t + 3) = 0
Setting each factor equal to zero, we have two possible solutions:
t - 6 = 0 => t = 6
t + 3 = 0 => t = -3
Therefore, the solutions to the equation t^2 - 3t - 18 = 0 by factoring are t = 6 and t = -3.
2. Completing the Square:
To complete the square, we will add and subtract the square of half the coefficient of the linear term to the equation.
t^2 - 3t - 18 = 0
t^2 - 3t = 18
(t^2 - 3t + 9/4) = 18 + 9/4 (add (3/2)^2 = 9/4 on both sides)
(t - 3/2)^2 = 81/4
Taking the square root of both sides:
t - 3/2 = ±√(81/4)
t - 3/2 = ±9/2
t = 3/2 ±9/2
t = (3 + 9)/2 or t = (3 - 9)/2
t = 12/2 or t = -6/2
t = 6 or t = -3
Therefore, the solutions to the equation t^2 - 3t - 18 = 0 by completing the square are t = 6 and t = -3, same as the factoring method.
3. Quadratic Formula:
The quadratic formula is given by t = (-b ± √(b^2 - 4ac)) / (2a) for a quadratic equation in the form of at^2 + bt + c = 0.
From the equation t^2 - 3t - 18 = 0, we have a = 1, b = -3, and c = -18.
t = (-(-3) ± √((-3)^2 - 4(1)(-18))) / (2(1))
t = (3 ± √(9 + 72)) / 2
t = (3 ± √(81)) / 2
t = (3 ± 9) / 2
So, t = (3 + 9) / 2 or t = (3 - 9) / 2
t = 12 / 2 or t = -6 / 2
t = 6 or t = -3
Again, we obtain the same solutions as the factoring and completing the square methods.
In conclusion, the solutions to the equation t^2 - 3t - 18 = 0 using factoring, completing the square, and the quadratic formula are t = 6 and t = -3.
To solve the quadratic equation t^2 - 3t - 18 = 0 by factoring, completing the square, and using the quadratic formula, let's go through each method step by step.
Method 1: Factoring
To factor the quadratic equation, we are looking for two numbers that multiply to -18 and add up to -3. The numbers -6 and 3 satisfy these conditions, so we can factor the equation as follows:
(t - 6)(t + 3) = 0
Setting each factor equal to zero, we get:
t - 6 = 0 or t + 3 = 0
Solving these equations gives us:
t = 6 or t = -3
Method 2: Completing the Square
To complete the square, we first need to rewrite the equation in the form (t - h)^2 = k.
t^2 - 3t - 18 = 0
Rearrange the equation by moving the constant term to the right side:
t^2 - 3t = 18
To complete the square, take half of the coefficient of t (-3/2), square it ((-3/2)^2 = 9/4), and add it to both sides:
t^2 - 3t + 9/4 = 18 + 9/4
Simplify the right side:
t^2 - 3t + 9/4 = 81/4
Now, rewrite the left side as a perfect square:
(t - 3/2)^2 = 81/4
Take the square root of both sides:
t - 3/2 = ±√(81/4)
Simplify:
t - 3/2 = ±9/2
Add 3/2 to both sides:
t = 3/2 ±9/2
This gives us the solutions:
t = 6 or t = -3
Method 3: Quadratic Formula
The quadratic formula is given by:
t = (-b ± √(b^2 - 4ac)) / (2a)
For the quadratic equation t^2 - 3t - 18 = 0, the coefficients are a = 1, b = -3, and c = -18. Substituting these values into the formula, we have:
t = (-(-3) ± √((-3)^2 - 4(1)(-18))) / (2(1))
Simplifying:
t = (3 ± √(9 + 72)) / 2
t = (3 ± √81) / 2
t = (3 ± 9) / 2
This gives us the solutions:
t = 6 or t = -3
Comparing the results:
All three methods give the same solutions: t = 6 and t = -3. This confirms the correctness of our calculations.
To solve the quadratic equation t^2 - 3t - 18 = 0, we can use different methods: factoring, completing the square, and the quadratic formula. Here's the step-by-step process for each method:
1. Factoring:
Step 1: Rewrite the equation: t^2 - 6t + 3t - 18 = 0
Step 2: Group the terms and factor by grouping: (t^2 - 6t) + (3t - 18) = 0
Step 3: Factor out common terms from each group: t(t - 6) + 3(t - 6) = 0
Step 4: Notice that we have a common factor of (t - 6): (t - 6)(t + 3) = 0
Step 5: Equate each factor to zero and solve for t:
t - 6 = 0 -> t = 6
t + 3 = 0 -> t = -3
The factored form of the equation is (t - 6)(t + 3) = 0, and the solutions are t = 6 and t = -3.
2. Completing the Square:
Step 1: Rewrite the equation: t^2 - 3t - 18 = 0
Step 2: Move the constant term to the other side: t^2 - 3t = 18
Step 3: Add the square of half the coefficient of the x-term to both sides of the equation. The coefficient of the t-term is -3, so half of it is -3/2, and the square of -3/2 is 9/4:
t^2 - 3t + 9/4 = 18 + 9/4
Step 4: Simplify both sides of the equation: (t - 3/2)^2 = 81/4
Step 5: Take the square root of both sides: t - 3/2 = ± sqrt(81/4)
Step 6: Simplify the right side: t - 3/2 = ± (9/2)
Step 7: Solve for t by considering both possibilities:
t - 3/2 = 9/2 -> t = 9/2 + 3/2 -> t = 12/2 -> t = 6
t - 3/2 = -9/2 -> t = -9/2 + 3/2 -> t = -6/2 -> t = -3
The solutions obtained using the completing the square method are t = 6 and t = -3.
3. Quadratic Formula:
The quadratic formula is given by the formula: t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In the equation t^2 - 3t - 18 = 0, the coefficients are:
a = 1
b = -3
c = -18
Plugging these values into the quadratic formula, we get:
t = (-(-3) ± sqrt((-3)^2 - 4(1)(-18))) / (2(1))
t = (3 ± sqrt(9 + 72)) / 2
t = (3 ± sqrt(81)) / 2
t = (3 ± 9) / 2
Solving for t, we have two possibilities:
(t + 6) / 2 = 9/2 -> t + 6 = 9 -> t = 9 - 6 -> t = 3/2
(t - 6) / 2 = -9/2 -> t - 6 = -9 -> t = -9 + 6 -> t = -3
The solutions obtained using the quadratic formula are t = 3/2 and t = -3.
Comparing the results obtained from all three methods:
- Factoring: t = 6 and t = -3
- Completing the square: t = 6 and t = -3
- Quadratic formula: t = 3/2 and t = -3
As you can see, all three methods yield the same solutions for the given quadratic equation.