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solving quadratic with completing the square

1a) 4x^2 - 24x -189 in form a(x+b)^2 +c
which i got as 4(x-3)^2 - 225

b) using answer to part a, solve equation
4x^2 -24x - 189 = 0

what do i do with 4(x-3)^2 - 225 ?
Reply 1
Avatar for Zacken
Zacken
22
Well you know that you can write the LHS in the completed the square form, so 4(x−3)2−225=04\left(x-3\right)^2 - 225= 0 , manipulate:

4(x−3)2=225⇒(x−3)2=2254⇒x−3=±2254⇒⋯\displaystyle 4(x-3)^2 = 225 \Rightarrow (x-3)^2 = \frac{225}{4}\Rightarrow x -3 = \pm \sqrt{\frac{225}{4}} \Rightarrow \cdots
Have you tried making it equal to 0 and solving for x?
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Reply 3
Avatar for Aph
Aph
22
Rearrange for X....

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Reply 4
Avatar for TSR George
TSR George
OP
17
Original post by Zacken
Well you know that you can write the LHS in the completed the square form, so 4(x−3)2−225=04\left(x-3\right)^2 - 225= 0 , manipulate:

4(x−3)2=225⇒(x−3)2=2254⇒x−3=±2254⇒⋯\displaystyle 4(x-3)^2 = 225 \Rightarrow (x-3)^2 = \frac{225}{4}\Rightarrow x -3 = \pm \sqrt{\frac{225}{4}} \Rightarrow \cdots


Original post by Matrix123
Have you tried making it equal to 0 and solving for x?
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Original post by Aph
Rearrange for X....

Posted from TSR Mobile


oh ok i over thought it, thank you
Fair enough :smile: I over think all the time tbh :tongue:
You're welcome :smile:
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