My Math Forum The Discriminant

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August 13th, 2017, 07:09 AM   #1
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The Discriminant

Firstly would you would say that this comes under Linear or Abstract Algebra? If I've posted it in the wrong place sorry.
I have been confuddled by this question. Find the range of values of a for which the following equations have two real roots: ax^2 + 2x +3 = 0
I've given it a couple of goes, but can't seem to get the correct answer. See attached for my scruffy workings.
If someone could just show me how they would work through the problem then I will probably be able to see where I am doing wrong.

Many thanks
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August 13th, 2017, 07:15 AM   #2
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Quote:
 Originally Posted by EvolvedPie Firstly would you would say that this comes under Linear or Abstract Algebra?
Neither. This is high school math, not college.

Anyway, I don't see how you get from $\Delta = 2^2 - 4*a*3$ to
$\Delta = 4 - 12a - 12$

August 13th, 2017, 07:27 AM   #3
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Hi Micrm@ss
Quote:
 Originally Posted by Micrm@ss Neither. This is high school math, not college.
Yes I realised that after posting. I'm from the UK so we use the wrong words for things.

And thank you. You may not realise it but you just pointed out to me what I was doing wrong.
b^2-4ac
I was putting 4(a*c) meaning that everything was being multiplied by 4 as opposed to just multiplying them together.
User error. Thanks again.

 August 14th, 2017, 07:24 AM #4 Newbie     Joined: Aug 2017 From: Bath, UK Posts: 4 Thanks: 0 Right, after realising what I was doing wrong I could get the correct answers for a couple of questions. But not this one: X2 + 8x + a =0 D = 8^2 – 4 * 8 * a = 0 D= 64 – 32a = 0 Divide by 32: D = 2 – a = 0 Which would mean that the answer is a < 2 But the answer is apparently a < 16. What am I doing wrong? Many thanks
August 14th, 2017, 07:46 AM   #5
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Quote:
 X2 + 8x + a =0 D = 8^2 – 4 * 8 * a = 0
$y = ax^2 + bx + c \implies D = b^2 - 4ac$

for $y = x^2 + 8x + a$ ... note that $a$ is the constant $c$ and $1$ is the leading coefficient, $a$, in the above equation for the discriminant, $D$

suffice to say, $a$ was a poor choice to represent the constant term.

$D = 8^2 - 4 \cdot 1 \cdot a = 64-4a = 0 \implies a = 16$

 August 14th, 2017, 07:48 AM #6 Newbie     Joined: Aug 2017 From: Bath, UK Posts: 4 Thanks: 0 Thanks Skeeter.
August 14th, 2017, 07:55 AM   #7
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Quote:
 Originally Posted by EvolvedPie I'm from the UK so we use the wrong words for things.
No we don't! Although it's perfectly understandable at your level not to know what Linear Algebra and Abstract Algebra are or that quadratic equations are neither. Don't worry about it.

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