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completing the sqare

i was revising and couldnt do this equation, the asnwer was a= 6, b= 2, c= -19, i understand how to get a and b but cant seem to figure out c, the equation was the following:
6x^2 + 24x + 5 = a(x+b)^2 + c

help? x
Original post by hinalllx
i was revising and couldnt do this equation, the asnwer was a= 6, b= 2, c= -19, i understand how to get a and b but cant seem to figure out c, the equation was the following:
6x^2 + 24x + 5 = a(x+b)^2 + c

help? x


Sure, gimme a few mins and I'll write up an answer for ya! :smile:
Original post by Mehru1214
Sure, gimme a few mins and I'll write up an answer for ya! :smile:


6x^2 + 24x + 5 = 0 First factorise the ax^2 + bx bit

6[x^2 + 4x] + 5 = 0 Now complete the square in the brackets.

6[(x+2)^2 - 4] + 5 = 0 Expand the square brackets.

6(x+2)^2 - 24 + 5 = 0 Now simplify!

6(x+2)^2 - 19 = 0 DONE!!!

a = 6
b = 2
c = -19
Original post by Mehru1214
6x^2 + 24x + 5 = 0 First factorise the ax^2 + bx bit

6[x^2 + 4x] + 5 = 0 Now complete the square in the brackets.

6[(x+2)^2 - 4] + 5 = 0 Expand the square brackets.

6(x+2)^2 - 24 + 5 = 0 Now simplify!

6(x+2)^2 - 19 = 0 DONE!!!

a = 6
b = 2
c = -19


Please don’t post full solutions in the maths forums
Reply 4
Original post by Mehru1214
6x^2 + 24x + 5 = 0 First factorise the ax^2 + bx bit

6[x^2 + 4x] + 5 = 0 Now complete the square in the brackets.

6[(x+2)^2 - 4] + 5 = 0 Expand the square brackets.

6(x+2)^2 - 24 + 5 = 0 Now simplify!

6(x+2)^2 - 19 = 0 DONE!!!

a = 6
b = 2
c = -19


thank you!! x

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