# Thread: How to find "a" in this quadratic equation that is set to zero?

1. ## How to find "a" in this quadratic equation that is set to zero?

The quadratic equation x^2 + ax + 1 = 0 might seem easy to solve for x, but I have zero clue when it comes to solving it for a!
Could someone give a heads up on how to solve for a so that the equation has one solution? Thanks!

2. ## Re: How to find "a" in this quadratic equation that is set to zero?

Perhaps you could write out the question. If $x$ is a variable, it doesn't really make sense to simply solve for the constant $a$. That said, rearranging the equation so that $a$ is the subject shouldn't present great difficulties: move the terms that don't contain $a$ to the other side and then divide by whatever multiplies the $a$.

3. ## Re: How to find "a" in this quadratic equation that is set to zero?

Originally Posted by bossbasslol
The quadratic equation x^2 + ax + 1 = 0 might seem easy to solve for x, but I have zero clue when it comes to solving it for a!
Could someone give a heads up on how to solve for a so that the equation has one solution? Thanks!
I think you are trying to find values of a such that the quadratic has a double root. This means the discriminant is zero.

Here we have $\displaystyle D = a^2 - 4 \cdot 1 \cdot 1$. For what values of a is D = 0?

-Dan

4. ## Re: How to find "a" in this quadratic equation that is set to zero?

quadratic formula

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

$x$ has a single solution when $\sqrt{a^2-4}=0$

i.e. what Topsquark said

5. ## Re: How to find "a" in this quadratic equation that is set to zero?

Originally Posted by bossbasslol
The quadratic equation x^2 + bx + 1 = 0 might seem easy to solve for x, but I have zero clue when it comes to solving it for b!
Could someone give a heads up on how to solve for a so that the equation has one solution?
Traditionally the quadratic equation is $ax^2+bx+c=0$ and has roots $\dfrac{-b\pm\sqrt{b^2-4(a)(c)}}{2a}$ (Please note that your question does not use standard notation).
The expression $\delta=b^2-4(a)(c)$ is known as the discriminate. That is due to the fact that the value of $\delta$ discriminates among the kinds of roots.
$\delta~\begin{cases}>0 &: \text{ two real roots } \\ =0 &: \text{ one real roots} \\<0 &: \text{ no real roots }\end{cases}$
Changing to standing to standard notation $x^2+bx+1=0$ has one real root if $\delta=b^2-4(1)(1)=0$.

6. ## Re: How to find "a" in this quadratic equation that is set to zero?

So from what I am absorbing, in x^2 + ax +1 the two roots (the one solution) must be found by inserting the respective a, b, and c values in the quadratic formula.
This means that x^2 houses a=1, ax contains b=a, and 1 is just one. Right?

7. ## Re: How to find "a" in this quadratic equation that is set to zero?

Originally Posted by bossbasslol
So from what I am absorbing, in x^2 + ax +1 the two roots (the one solution) must be found by inserting the respective a, b, and c values in the quadratic formula.
This means that x^2 houses a=1, ax contains b=a, and 1 is just one. Right?
I think you meant to type c=1, which is correct

8. ## Re: How to find "a" in this quadratic equation that is set to zero?

Thank you, and yes, I meant c=1.

This x, , it must be plugged into x^2 + ax + 1 to find a, right? Or why would it be set equal to zero? A little confused here.

9. ## Re: How to find "a" in this quadratic equation that is set to zero?

Originally Posted by bossbasslol
So from what I am absorbing, in x^2 + ax +1 the two roots (the one solution) must be found by inserting the respective a, b, and c values in the quadratic formula. This means that x^2 houses a=1, ax contains b=a, and 1 is just one. Right?
I don't know who wrote this question. In any case I understand that you are not in any way responsible for this.
So lets say it is $x^2+kx+1=0$ what value of $k$ gives one real root.
Well the discriminate is $k^2-4(1)(1)$ which is zero of $k=\pm 2$
Again, you are not at fault an any way that whoever wrote this question either deliberately or stupidly mislead readers.

10. ## Re: How to find "a" in this quadratic equation that is set to zero?

Originally Posted by bossbasslol
Thank you, and yes, I meant c=1.

This x, , it must be plugged into x^2 + ax + 1 to find a, right? Or why would it be set equal to zero? A little confused here.
No.

You said that you want to find the value of $a$ such that there is a single root to $x^2+ax+1=0$

We've all told you that in order for that to be so the discriminate $D = a^2 -4 = 0$

It should be pretty clear that $a=\pm 2$

so the polynomials

$x^2 + 2x + 1$ and $x^2-2x+1$ both have only a single root.

This isn't surprising as they are recognized as $(x+2)^2$ and $(x-2)^2$

11. ## Re: How to find "a" in this quadratic equation that is set to zero?

Yes, thank you for clearing things up for me. Something was off in that question, and it was made concise with the retrieval of missing pieces.