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!
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!
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 $a$.
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 $\delta=b^2-4(1)(1)=0$.
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.
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 $(x-2)^2$