P(x)=ax^2+bx+c

**yoman360** · Jun 4th 2009, 07:40 PM

True or False?
If a, b, and c denote real numbers and p(x)=ax^2+bx+c, x € ℝ, then there exists a real number x0 such that p(x0)=0.

Please show work

**Amer** · Jun 4th 2009, 07:58 PM

Originally Posted by yoman360

True or False?
If a, b, and c denote real numbers and p(x)=ax2+bx+c, x € ℝ, then there exists a real number x0 such that p(x0)=0.

Please show work

if

$b^2-4ac>0$

then there exist two different solutions

if

$b^2-4ac<0$

there is no solutions

$b^2-4ac=0$

there are two same solutions
for more information see link below
Quadratic function - Wikipedia, the free encyclopedia

**VonNemo19** · Jun 4th 2009, 08:31 PM

Originally Posted by yoman360

True or False?
If a, b, and c denote real numbers and p(x)=ax2+bx+c, x € ℝ, then there exists a real number x0 such that p(x0)=0.

Please show work

False.

Putting this equation in standard form we have

$p(x)=ax^2+bx+c$
$=a(x^2+\frac{b}{a}x)+c$
$=a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2})+c$
$=a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2})-\frac{b^2}{4a}+c$
$=a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c$

Assume that $a>0$ , we can conclud that this is graph of a parabola that opens upwards and has vetex $(-\frac{b}{2a},c-\frac{b^2}{4a})$ . Again, since the parabola opens upward, $p(x_0)=0\Longleftrightarrow{c-\frac{b^2}{4a}\leq0}$

And since there are enumerable instances where this statement does not hold, the answer is false

**yoman360** · Jun 13th 2009, 07:45 PM

Originally Posted by VonNemo19

False.

Putting this equation in standard form we have

$p(x)=ax^2+bx+c$
$=a(x^2+\frac{b}{a}x)+c$
$=a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2})+c$
$=a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2})-\frac{b^2}{4a}+c$
$=a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c$

Assume that $a>0$ , we can conclud that this is graph of a parabola that opens upwards and has vetex $(-\frac{b}{2a},c-\frac{b^2}{4a})$ . Again, since the parabola opens upward, $p(x_0)=0\Longleftrightarrow(-\frac{b}{2a},c-\frac{b^2}{4a})\leq0$

And since there are enumerable instances where this statement does not hold, the answer is false

what if a<0 then it would cross the x-axis thus making the statement true.

**Chris L T521** · Jun 13th 2009, 09:23 PM

Originally Posted by VonNemo19

False.

...

Assume that $a>0$ , we can conclude that this is graph of a parabola that opens upwards and has vertex $(-\frac{b}{2a},c-\frac{b^2}{4a})$ . Again, since the parabola opens upward, $p(x_0)=0\Longleftrightarrow{\color{red}(-\frac{b}{2a},c-\frac{b^2}{4a})\leq0}$

The part in red doesn't make sense. What I think you're trying to say is that $p\!\left(x_0\right)=0\iff c-\frac{b^2}{4a}\leq0$ . An ordered pair cannot be less that a number.

And since there are enumerable instances where this statement does not hold, the answer is false

But the question isn't asking if this holds for all cases. The statement is an existential statement; one that asks us to find at least one case where the statement is true.

True or False?
If $a,b,c\in\mathbb{R}$ and $p\!\left(x\right)=ax^2+bx+c$ , $x\in\mathbb{R}$ , then there exists a real number $x_0$ such that $p\!\left(x_0\right)=0$ .

This is a true statement under certain cases. Again, it all depends on the value of the discriminant $\Delta = b^2-4ac$ . If

$\Delta > 0$ , then there are two solutions (so there exists more than one $x_0$ ).

$\Delta = 0$ , there there is only one solution (thus $x_0$ exists).

$\Delta < 0$ , there there is no real solution. This implies that a $x_0$ does not exist.

To summarize, the only time when an $x_0$ exists is when $\Delta\geq 0$ . Also, the value of $x_0$ is the real zero(s) of the quadratic equation.

Consider $f\!\left(x\right)=x^2+3x+2$ . Since $1,3,2\in\mathbb{R}$ , we search for an $x_0$ such that $f\!\left(x_0\right)=0$ . So in essence, we solve $x_0^2+3x_0+2=0$ . But $x_0^2+3x_0+2=0\implies\left(x_0+2\right)\left(x_0+ 1\right)=0\implies\left(x_0+2\right)=0$ or $\left(x_0+1\right)=0$ . So in this case, there exists two possible values for $x_0$ : $x_0=-2$ and $x_0=-1$ , where both values for $x_0$ are real numbers.

**yoman360** · Jun 13th 2009, 09:48 PM

Originally Posted by Chris L T521

The part in red doesn't make sense. What I think you're trying to say is that $p\!\left(x_0\right)=0\iff c-\frac{b^2}{4a}\leq0$ . An ordered pair cannot be less that a number.

But the question isn't asking if this holds for all cases. The statement is an existential statement; one that asks us to find at least one case where the statement is true.

This is a true statement under certain cases. Again, it all depends on the value of the discriminant $\Delta = b^2-4ac$ . If

$\Delta > 0$ , then there are two solutions (so there exists more than one $x_0$ ).

$\Delta = 0$ , there there is only one solution (thus $x_0$ exists).

$\Delta < 0$ , there there is no real solution. This implies that a $x_0$ does not exist.

To summarize, the only time when an $x_0$ exists is when $\Delta\geq 0$ . Also, the value of $x_0$ is the real zero(s) of the quadratic equation.

Consider $f\!\left(x\right)=x^2+3x+2$ . Since $1,3,2\in\mathbb{R}$ , we search for an $x_0$ such that $f\!\left(x_0\right)=0$ . So in essence, we solve $x_0^2+3x_0+2=0$ . But $x_0^2+3x_0+2=0\implies\left(x_0+2\right)\left(x_0+ 1\right)=0\implies\left(x_0+2\right)=0$ or $\left(x_0+1\right)=0$ . So in this case, there exists two possible values for $x_0$ : $x_0=-2$ and $x_0=-1$ , where both values for $x_0$ are real numbers.

Oh ok I get it now

can you help me with this one:
http://www.mathhelpforum.com/math-he...z-1-2-1-a.html

**yoman360** · Jun 13th 2009, 09:57 PM

Wouldn't the answer be true? because 0 is a real number and if b and c= 0 then then $p(x_0)=0$
$p(x)= ax^2 +bx +c$
$p(x) = 1x^2+0x+0$
$p(x) = x^2$
then there's only one solution of the function.

**Chris L T521** · Jun 13th 2009, 10:07 PM

Originally Posted by yoman360

Wouldn't the answer be true? because 0 is a real number and if b and c= 0 then then $p(x_0)=0$
$p(x)= ax^2 +bx +c$
$p(x) = 1x^2+0x+0$
$p(x) = x^2$
then there's only one solution of the function.

Yes, that is true. Note that you imposed certain conditions on what you let a,b, and c equal. As I said above, the statement is true for certain cases/conditions, and this would be one of them. You just illustrated the case where we only have one value $x_0$ such that $p\!\left(x_0\right)=0$ (I illustrated a case where we had more than one value).

**VonNemo19** · Jun 14th 2009, 10:14 AM

Originally Posted by Chris L T521

The part in red doesn't make sense. What I think you're trying to say is that $p\!\left(x_0\right)=0\iff c-\frac{b^2}{4a}\leq0$ . An ordered pair cannot be less that a number.

But the question isn't asking if this holds for all cases. The statement is an existential statement; one that asks us to find at least one case where the statement is true.

This is a true statement under certain cases. Again, it all depends on the value of the discriminant $\Delta = b^2-4ac$ . If

$\Delta > 0$ , then there are two solutions (so there exists more than one $x_0$ ).

$\Delta = 0$ , there there is only one solution (thus $x_0$ exists).

$\Delta < 0$ , there there is no real solution. This implies that a $x_0$ does not exist.

To summarize, the only time when an $x_0$ exists is when $\Delta\geq 0$ . Also, the value of $x_0$ is the real zero(s) of the quadratic equation.

Consider $f\!\left(x\right)=x^2+3x+2$ . Since $1,3,2\in\mathbb{R}$ , we search for an $x_0$ such that $f\!\left(x_0\right)=0$ . So in essence, we solve $x_0^2+3x_0+2=0$ . But $x_0^2+3x_0+2=0\implies\left(x_0+2\right)\left(x_0+ 1\right)=0\implies\left(x_0+2\right)=0$ or $\left(x_0+1\right)=0$ . So in this case, there exists two possible values for $x_0$ : $x_0=-2$ and $x_0=-1$ , where both values for $x_0$ are real numbers.

So, you maintain that there exists a zero for every parabola in the form of $ax^2+b^2+c$ ?

You were right about the last statement of my argument, a small error, but the result is still sound. I cahnged it. But by reading the question again, and taking note of the phrase "there exists" , I interpret this to mean that "there must exist" or 'there always exists". And you and I both know that this is a false statement.

I will attempt to reword the statement. Note that a,b, and c can be any real numbers.

--For any given parabola, there exists at least one place where it touches the x-axis.--

Is this ridiculous, or is this ridiculous?

"Exists is not a predicate."Immanuel Kant.

**HallsofIvy** · Jun 14th 2009, 11:25 AM

The original statement was, essentially that "there exist a real number solution to every quadratic equation". The simplest way to show that is false is to take a= 1, b= 0, c= 1. There is no real number, x, such that $x^2+ 1= 0$ .

**yoman360** · Jun 14th 2009, 12:48 PM

Ok it makes sense now since the statement is not always true its false.

**VonNemo19** · Jun 14th 2009, 12:55 PM

Originally Posted by yoman360

Ok it makes sense now since the statement is not always true its false.

Indubidubly!

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