What is the value of $t$?

Question

$$x^2 -tx +9=0 $$ The second root of this equation is given as $1$. What is the value of $t$?

Unfortunalety, I'm stuck at this problem. I'm getting $t = \pm 10$ for $x = 1$ However, the right answer seems $\frac {25}{4}$

UPDATE: The second root (which is $x$, was given as $\frac{1}{\sqrt x} + \sqrt x = 2$

Answer 1

It seems to be 10, indeed let condider

$$x^2-tx+9=(x-a)(x-b)=x^2-(a+b)x+ab\implies ab=9$$

and if $a=1\implies b=9$ and thus $t=a+b=10$.

Answer 2

Performing the quadratic formula:

$$x^2-tx+9=0 \implies x=\dfrac{t\pm\sqrt{t^2-36}}2$$

Since one root is $1$:

$$\dfrac{t+\sqrt{t^2-36}}2=1$$

From which you get $t=10$, but $t=10$ is extraneous.

or

$$\dfrac{t-\sqrt{t^2-36}}2=1$$

From which $t=10$.

$\dfrac {25}4$ is incorrect.

So, $t=10$ is correct.
To check, $t^2-10t+9=0 \implies (t-9)(t-1)=0$, which has a root at $x=1$.

Answer 3

If $f(x) = x^2−tx+9$ has a root at $x=1$, then that means that $f(1) = 0$.

In particular, we find that $1^2 - t + 9 = 0$, i.e., $t=10$.

Answer 4

In fact, the answer is $10$, since $x=1$. It was given that one of roots for $x$ is $1$, which leads us to quadratic solutions:

$$\dfrac{t+\sqrt{t^2-36}}2=1$$ or $$\dfrac{t-\sqrt{t^2-36}}2=1$$ leading to $\sqrt{t^2-36}=2-t$ or $\sqrt{t^2-36}=t-2$. Both equations lead to one equation of $t$: $$t^2-36=t^2-4t+4$$ giving $t=10$ as the only answer, since $t^2$ terms canceled out.

For interesting readers: Also note that, extended work would show you that $\frac{25}{4}$ would be a solution for $t$ only if $x=4$. Further, $t$ also has integer solutions when $x=3$ $(t=6)$ and $x=9$ $(t=10)$.