1. Solve by completing square

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

3. Solve with quadratic formula

1. Solve by completing square

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

3. Solve with quadratic formula

Tutors, please sign in to answer this question.

San Diego, CA

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

Olney, MD

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

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 - 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

4(r

r

r

(r - (7/2))

r - 7/2 = ±√(98/4) = ±(7/2)√2

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

Thankyou for taking your time out to help me it means a lot!

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