Since Surendra answered the first question, I'll answer the next two. Let's put the equation into standard form:
4r^{2} - 28r - 49 = 0
2. The discriminant is b^{2} - 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)√(b^{2}-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):
4r^{2} - 28r = 49
4(r^{2} - 7r) = 49
r^{2} - 7r = 49/4
r^{2} - 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)
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