how do i solve 7r^2+48r=64

Question

This equation i need to solve by factoring but i don't remember how to do it can someone help?

Answer 1

solve 7r^2+48r=64

 

Get zero on one side of the equals sign by subtracting 64 from both sides.

7r^2+48r  - 64 =64  - 64

solve 7r^2+48r - 64 = 0

Factor the left side into a product of two binomials where the first term in each binomial multiplies to the first term in the quadratic on the left side and the last terms in each binomial multiplies to the last term of the quadratic.

(7r +   $ )(r  + &) = 0  (7 is prime so it only has factors of itself and 1)

to find the last terms (7 x &) + (1 x $) = 48   (& x $ should equal 64)

-64 = (-2 x 32) or (2 x -32) or (-4 x 16) or (4 x -16) or (-8 x 8)

 

Try each factor in (7 x &) + (1 x $) = 48    (-8 and 8 work)

(7r + - 8)(r + 8) = 0 

 

zero product rule says one of the binomials should equal zero so..

(7r - 8) = 0  or (r + 8) = 0

 

Solve each..

7r - 8 = 0 --> 7r = 8 --> r = 8/7

or 

r + 8 = 0 --> r = -8 

 

r = 8/7 or -8

 

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Answer 2

7r² + 48r - 64 = 0      -64*7 = -448    You want 2 factors of -448 that add up to +48.  One will be + and the other -. They would be +56, -8. Split the middle term to incorporate these 2 factors:

 

7r² + 56r - 8r - 64 = 0   Now split the 4 terms into 2 pairs and factor:

 

7r(r + 8) - 8(r + 8)   Combine:

 

(7r - 8)(r + 8) ANSWER

 

Answer 3

7r^2+48r=64

7r^2+48r-64=0

multiply 7*64 in factored form

7*2*2*2*2*2*2

now group the factors so that the two factors subtract to make 48

(7*2*2*2)*(2*2*2)=56*8 and 56-8=48

now factor

(7r     )(r     )=0 so far

(7r   8)(r   8)=0 so far

48 is positive so 56 must be positive and 8 must be negative

(7r-8)(r+8)=0

set each factor to zero and solve

7r-8=0

7r=8

r=8/7

r+8=0

r=-8