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Can you solve this?

Solve 2x^2+x-351=0
By the way, the only way I want this presented is by completing the square.
(edited 7 years ago)
Is this a way of getting us to do your homework?
Reply 2
Avatar for fefssdf
fefssdf
20
Original post by Ioniccopperflame
Solve 2x^2+x+351=0
By the way, the only way I want this presented is by completing the square.


divide through by 2 and then do completing the square as usual ?
Original post by Ioniccopperflame
Solve 2x^2+x+351=0
By the way, the only way I want this presented is by completing the square.


Yes, I can solve it. I would, however, prefer for you to try first.

Just so you know, both roots are complex, so they are far beyond GCSE understanding.
Original post by TheOtherSide.
Is this a way of getting us to do your homework?


No I already figured out the answer. It's x=-13.5 or x=13. But whenever I do the same question by completing the square, I get 1/2 +-sqrt175.75. So I just wanted to know if I've made an error or not.
(edited 7 years ago)
Original post by Ioniccopperflame
No I already figured out the answer. It's x=-13.5 or x=13. But whenever I do the same question by completing the square, I get 1/2 +-sqrt175.75. So I just wanted to know if I've made an error or not.


You shouldn't be getting real solutions, unless you've made an error in copying the question.
Original post by Ioniccopperflame
No I already figured out the answer. It's x=-13.5 or x=13. But whenever I do the same question by completing the square, I get 1/2 +-sqrt175.75. So I just wanted to know if I've made an error or not.

Neither of those is a solution to the equation that you posted - was there an error in it? Should it have been -351?
(edited 7 years ago)
Original post by Ioniccopperflame
Solve 2x^2+x+351=0
By the way, the only way I want this presented is by completing the square.


x=+- 69, for all real and imaginary values of x in da world :O
Original post by _gcx
Yes, I can solve it. I would, however, prefer for you to try first.

Just so you know, both roots are complex, so they are far beyond GCSE understanding.


No I already figured out the answer. It's x=-13.5 or x=13. But whenever I do the same question by completing the square, I get 1/2 +-sqrt175.75. So I just wanted to know if I've made an error or not when I tried solving it through completing the square.
Original post by Ioniccopperflame
No I already figured out the answer. It's x=-13.5 or x=13. But whenever I do the same question by completing the square, I get 1/2 +-sqrt175.75. So I just wanted to know if I've made an error or not when I tried solving it through completing the square.


Both of those are wrong. This equations has no real roots.
Original post by Ioniccopperflame
Solve 2x^2+x-351=0
By the way, the only way I want this presented is by completing the square.

I'm so sorry. I realised I made a mistake in the forum. I meant to type -351 instead of +351 but it's changed now
Original post by Ioniccopperflame
Solve 2x^2+x-351=0
By the way, the only way I want this presented is by completing the square.


I'm so sorry. I realised I made a mistake in the forum. I meant to put -351 instead of +351. I've changed it now.
Original post by Ioniccopperflame
Solve 2x^2+x-351=0
By the way, the only way I want this presented is by completing the square.


I'll try 2x^2+x+351 for fun anyway
Reply 13
Avatar for B_9710
B_9710
18
x2+12x−3512=0 \displaystyle x^2+\frac{1}{2}x-\frac{351}{2}=0
(x+14)2−116−280816=0 \displaystyle \left (x+\frac{1}{4}\right )^2-\frac{1}{16}-\frac{2808}{16}=0
(x+14)2=280916 \displaystyle \left (x+\frac{1}{4}\right )^2 =\frac{2809}{16} .
I'm assuming you can do it from here.

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