+96 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.
+128475 (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
+128475 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
