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avatar+59 

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.

 Feb 10, 2020
 #1
avatar+109740 
+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

 

 

 

cool cool cool

 Feb 10, 2020
 #2
avatar+109740 
+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

 

 

cool cool cool

 Feb 10, 2020

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