Use 4r^2-28r=49 to answer problems

Question

1. Solve by completing square

2. Use the discriminant to find the number of unique real solutions

3. Solve with quadratic formula

Answer 1

4r^2  -  28r -49=0

 

4r^2 -28r + 49 -49 -49=0

 

4r^2 -28r + 49 = 98

 

(2r-7)^2 =  98

 

2r-7 = 9.899

 

2r = 16.899

 

r= 8.449

 

 

 

Answer 2

Since Surendra answered the first question, I'll answer the next two. Let's put the equation into standard form:

 

4r² - 28r - 49 = 0

 

2.  The discriminant is b² - 4ac  where a=4, b= -28, and c= -49.

 

(-28)² - (4)(4)(-49) = 784 + 784 = 1568

 

Since the discriminant is greater than zero, there will be two real solutions.  (The rules are: discriminant<0, no real solutions; discriminant=0, one solution; discriminant>0, two solutions).

 

3. Solve using the quadratic formula:

 

r = (-b/2a) ± (1/2a)√(b²-4ac)

Where a=4, b=-28, c=-49.

 

r = -(-28)/8 ± (1/8)√1568

r = 7/2 ± (7/2)√2

 

r =(7/2)(1+√2),  (7/2)(1-√2)   These are the exact solutions.  The approximate solutions are (rounded to two decimal places): r ≅ 8.45, -1.45

 

CHECK:

 4{(7/2)(1+√2)}² - 28{(7/2)(1+√2)} - 49 = 0

49 + 98√2 + 98 - 98 - 98√2 -49 = 0

49 - 49 = 0                                  

CHECK!

 

4{(7/2)(1-√2)}² - 28{(7/2)(1-√2)} - 49 = 0
49 - 98√2 + 98 - 98 + 98√2 - 49 = 0

49 - 49 = 0 

CHECK !

 

1. Here's another take on problem 1 (complete the square):

4r² - 28r = 49
4(r² - 7r) = 49
r² - 7r = 49/4
r² - 7r + (-7/2)² = 49/4 + (-7/2)²
(r - (7/2))² = 98/4
r - 7/2 = ±√(98/4) = ±(7/2)√2

r = (7/2) ± (7/2)√2= (7/2)(1±√2)