solve r^2-4r-32=0 by factoring, by completing the square, and by using the quadratic formula. Compare your results.
To solve the equation r^2 - 4r - 32 = 0 using factoring, we need to find two numbers whose product is -32 and whose sum is -4. The numbers -8 and 4 fit these requirements because (-8)(4) = -32 and -8 + 4 = -4. Therefore, we can factor the equation as (r - 8)(r + 4) = 0.
Setting each factor equal to 0 gives us two possible solutions: r - 8 = 0 or r + 4 = 0.
Solving these equations, we find r = 8 and r = -4.
To solve the equation by completing the square, we rewrite it in the form (r - b)^2 = c, where b is half the coefficient of r and c is a constant. In our equation, the coefficient of r is -4, so half of it is -2 (b = -2). To find c, we square b and add it to both sides of the equation:
r^2 - 4r - 32 = 0
(r - 2)^2 = 36
Taking the square root of both sides gives us:
r - 2 = ±6
r = 2 ±6
Solving for r, we find r = 8 and r = -4, which are the same solutions we obtained by factoring.
To solve the equation using the quadratic formula, the formula is r = (-b ±√(b^2 - 4ac)) / 2a. Plugging in the coefficients from our equation, a = 1, b = -4, and c = -32, we can substitute these values into the quadratic formula:
r = (-(-4) ±√((-4)^2 - 4(1)(-32))) / (2(1))
r = (4 ±√(16 + 128)) / 2
r = (4 ±√(144)) / 2
r = (4 ±12) / 2
r = 8 / 2 or -8 / 2
r = 4 or -4
Again, we find the same solutions as we obtained by factoring and completing the square.
Therefore, the solutions to the equation r^2 - 4r - 32 = 0 are r = 8 and r = -4.
To solve the quadratic equation r^2 - 4r - 32 = 0, let's use each of the three methods you mentioned: factoring, completing the square, and the quadratic formula.
1. Factoring the quadratic equation:
We look for two numbers that multiply to give -32 and add up to -4.
Factors of -32: -1, -2, -4, -8, 1, 2, 4, 8
Combination: -8 and 4
Now we can rewrite the equation as a product of two binomials:
(r - 8)(r + 4) = 0
Equating each factor to zero, we have two possible solutions:
r - 8 = 0 or r + 4 = 0
Solving for r in each case:
r = 8 or r = -4
So the solutions to the equation are r = 8 and r = -4.
2. Completing the square:
To complete the square, we add and subtract the square of half the coefficient of the middle term (which is -4/2 = -2) to the equation:
r^2 - 4r - 32 = 0
r^2 - 4r + (-2)^2 - (-2)^2 - 32 = 0
r^2 - 4r + 4 - 4 - 32 = 0
(r - 2)^2 - 36 = 0
Now, isolating the squared term:
(r - 2)^2 = 36
Taking the square root of both sides:
r - 2 = ±√(36)
r - 2 = ±6
Solving for r in each case:
r = 2 + 6 or r = 2 - 6
r = 8 or r = -4
The solutions are the same as obtained through factoring.
3. Quadratic formula:
The quadratic formula states that for the equation ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our quadratic equation r^2 - 4r - 32 = 0, a = 1, b = -4, and c = -32. Plugging these values into the quadratic formula:
r = (-(-4) ± √((-4)^2 - 4(1)(-32))) / (2(1))
r = (4 ± √(16 + 128)) / 2
r = (4 ± √(144)) / 2
r = (4 ± 12) / 2
Simplifying further:
r = (4 + 12) / 2 or r = (4 - 12) / 2
r = 16 / 2 or r = -8 / 2
r = 8 or r = -4
Again, the solutions are the same as obtained through factoring and completing the square.
In conclusion, the solutions to the quadratic equation r^2 - 4r - 32 = 0 are r = 8 and r = -4, and all three methods yield the same results.
To solve the quadratic equation r^2 - 4r - 32 = 0 using factoring, completing the square, and the quadratic formula, we will go through each method step by step.
1. Factoring:
To factor the quadratic equation, we need to find two numbers whose product is -32 and whose sum is -4 (coefficients of r^2 and r). The numbers that fit these criteria are -8 and 4 because -8 * 4 = -32 and -8 + 4 = -4. Therefore, we can rewrite the equation as:
(r - 8)(r + 4) = 0.
Setting each factor equal to zero gives us two solutions:
r - 8 = 0 → r = 8,
r + 4 = 0 → r = -4.
So, the solutions to the quadratic equation by factoring are r = 8 and r = -4.
2. Completing the square:
To complete the square, we will rearrange the equation by moving the constant term (-32) to the other side:
r^2 - 4r = 32.
Next, we take half of the coefficient of r(-4, in this case), square it, and add it to both sides of the equation:
r^2 - 4r + 4 = 32 + 4,
(r - 2)^2 = 36.
Now, taking the square root of both sides gives us:
r - 2 = ±√36,
r - 2 = ±6.
Solving for r gives us two solutions:
r = 2 + 6 = 8,
r = 2 - 6 = -4.
So, the solutions to the quadratic equation by completing the square are r = 8 and r = -4.
3. Quadratic Formula:
The quadratic formula for solving equations of the form ax^2 + bx + c = 0 is given by:
x = (-b ± √(b^2 - 4ac))/(2a).
In our equation, a = 1, b = -4, and c = -32. Plugging these values into the quadratic formula, we get:
r = (-(-4) ± √((-4)^2 - 4(1)(-32)))/(2(1)),
r = (4 ± √(16 + 128))/2,
r = (4 ± √(144))/2,
r = (4 ± 12)/2.
Simplifying further gives us two solutions:
r = (4 + 12)/2 = 16/2 = 8,
r = (4 - 12)/2 = -8/2 = -4.
Thus, the solutions to the quadratic equation using the quadratic formula are r = 8 and r = -4.
Comparing the results from all three methods, we can see that they all yield the same solutions, which are r = 8 and r = -4. Each method follows a different approach, but ultimately leads to the same outcome.