1)

For what values of j does the equation (2x+7)(x-5) = -43 + jx have exactly one real solution? Express your answer as a list of numbers, separated by commas.

2)

Find a such that ax^2+15x+4 is the square of a binomial.

daddypig Feb 10, 2020

#1**+1 **

(1)

(2x+7)(x-5) = -43 + jx simplify the left side

2x^2 + 7x -10x - 35 = -43 + jx rearrange as

2x^2 + (7 - j)x + 8 = 0

If this has one real solution.....then the discriminant must = 0

So

(7 -j)^2 - 4 (2)(8) =0

49 - 14j + j^2 - 64 = 0 rearrange as

j^2 -14j - 15 = 0 factor as

(j -15) ( j + 1) = 0

Set both facotrs to 0 and solve for j and we get that

j =15 and j = -1

So....we will have one real solution when j = -1 or j =15

See the graph here to confirm this : https://www.desmos.com/calculator/dw6lnjpnm7

CPhill Feb 10, 2020

#2**+1 **

2)

Find a such that ax^2+15x+4 is the square of a binomial.

For this to be true.....we need to have that the discrimant must = 0

So

15^2 - 4a (4) = 0

225 - 16a = 0

225 = 16a

a = 225 /16

If we take the positive square root of this we have 15/4

And note that (15/4 x + 2)^2 = (225/16)x^2 + 2 (15/4)(2)x + 4 = (225/16)x^2 + 15 x + 4

CPhill Feb 10, 2020