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

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2 Answers

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)

Comments

Hello Philip,
Yes, as per my methods,
I should have taken
2r-7= -9.899
r= -2.899
r= -1.4495
Hi Surrendra,
 
No problem.  I am the one who "liked" your answer since you posted a few minutes before I posted (not to mention that I made an error in my initial post).  I subsequently revised my answer because I realized there is an exact solution without rounding decimals.
 
v/r, Phil