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IF $a>0$ and $b<0$, which of the following statements are true about the value of (x) that solve the eq0ution $x^2 - ax + b = 0$

a)they have opposite signs b)their sum is greater than zero c)their product equals $- b$

Now my choice was a) and c), but c) is incorrect and I'm not sure why since b is negative (<0) which means that the values of x will be opposite (a negative product) hence a), but why not c)? The answer is a) and b) but I'm not sure why b) is an answer. It COULD be I think, but I believe c) has to be. Thanks everyone.

S.C.B.
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mike
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  • Remember that the sum of the roots is $a \gt 0$ and their product is $b \lt 0$. – dxiv Jan 29 '17 at 02:50
  • Ok if the product is b<0 then why is the answer choice not C which says -b? And what does the sum of the roots is a>0 mean? Thank you. – mike Jan 29 '17 at 02:57
  • The product is $(a^2 - (a^2-4b))/4= +b $. – fleablood Jan 29 '17 at 02:57
  • Sorry, what dos this mean? I'm not a math major, just studying for the GRE. I have no idea where your numbers came from. As a novice I know that the x are opposites, so the sum of them well depend on which number is negative and which is positive. – mike Jan 29 '17 at 03:01
  • @mike if the product is b<0 then why is the answer choice not C which says -b? Because the product is $b \lt 0,$, not $-b \gt 0,$. – dxiv Jan 29 '17 at 03:06
  • Because $b \ne -b $ unless $b=0$. If $b > 0$ then $b \ne -b $ and if $b < 0$ the $b \ne -b $. Zero is the only number that is equal to its negative. Whether b is positive or negative doesn't make any difference. – fleablood Jan 29 '17 at 03:17
  • Ok. But do you agree that the x values are opposites? One will be negative and one positive? And when we multiply them doesn't the product = the constant = b? A negative * a positive = - so -b? Obviously this is wrong, I just want to show my thought process for critique purposes. So I could write x^2 - x - 2 = 0 isn't the 2 the b and isn't this -2 = -b? If you decide to reply, could you please 'dumb' it down? Or dumb it down some more? Thank you. – mike Jan 29 '17 at 03:26
  • By the quadratic formula $x=\frac {a\pm\sqrt {a^2-4b}}2$ if you add them the discriminate cancels and you get a. Because b <a the discriminate is larger than a^2 so the one with subtracting will be less than zero. Multiplied you get the difference of squares which will give you b. b is not -b. It doesn't matter that b is negative. That just means -b is positive. Unless b is never equal to negative unless be is zero. – fleablood Jan 29 '17 at 03:28
  • Putting a negative sign in front of something does not make a negative number. It makes it the opposite sign. If $b $ is negative, then $-b $ is positive. – fleablood Jan 29 '17 at 03:32
  • @mike x^2 - x - 2 = 0 isn't the 2 the b and isn't this -2 = -b? No, that's $b=-2,$. Maybe it's more obvious if you work it backwards: take $x^2-ax+b$ then substitute $a=1$, $b=-2,$ and see what you get. – dxiv Jan 29 '17 at 04:28

1 Answers1

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$x^2-ax+b=0 \implies $

$x = \frac {a\pm\sqrt {a^2-4b}}2$

$b <0$ so $-4b>0$ so $\sqrt {a^2 -4b} >\sqrt {a^2}= a >0$.

So $a-\sqrt {a^2-4b} < a -a =0$.

So $\frac {a + \sqrt {a^2-4b} }2> 0 >\frac { a -\sqrt {a^2-4b}}2 $

so a)is true.

b) $\frac {a + \sqrt {a^2-4b}}2+ \frac {a - \sqrt {a^2-4b}}2 = a >0$ so b).

And $ \frac {a + \sqrt {a^2-4b}}2 \frac {a -\sqrt {a^2-4b}}2=\frac {a^2-(a^2-4b)}4=b \ne -b $ so not c)

fleablood
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  • Actually, since the solutions have opposite signs, their product is negative. And as b <0 then -b is positive. So the product can't be -b. You don't have to multiply them out. – fleablood Jan 29 '17 at 03:35
  • Ah, I think I see what you're saying. When you look at answer choice c) you're seeing -(b) and since we agree that the product is negative and the product is = b you're seeing -(-b)? I hope this is the case since that's what I was thinking COULD be the case with answer choice c). Basically answer choice c) is saying the answer is POSITIVE? – mike Jan 29 '17 at 03:37
  • No. I'm not seeing or saying anything of the sort! I am saying $-b \ne b $. It just doesn't. $b $ is a negative number and $-b $ is a positive number. The negative sign in front of a number has NOTHING to do with whether the number is negative or positive. $-b=(-1)*b $ and just as likely to be positive as it is to be negative. Remember if $k< 0$ then $|k|= -k$ because absolute values of a negative number is always positive and $-k$ is a positive number. Likewise $|-k|=-k$ because the absolute value of a positive number is a positive number. $-k $ is a POSITIVE number. – fleablood Jan 29 '17 at 07:14
  • great thanks for taking the time to help me out. – mike Feb 04 '17 at 21:40