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 $\frac{1}{\sqrt x} + \sqrt x = 2$

@XanderHenderson We can rewrite the equation as $x = \frac {t}[2} + \sqrt {\frac{t^2}[4}-9}$ — Goendo, Mar 03 '18 at 23:31
If the question is correct as posed, then I don't see why we can't just plug in x = 1 to get 1-t+9 = 0 $\implies$ t = 10. — James Yang, Mar 03 '18 at 23:32
@JamesYang Yes, that was what I've gotten. According to my textbook, that's wrong. — Goendo, Mar 03 '18 at 23:32
As $1^2 -(25/4) \cdot 1 + 9 \neq 0$, $25/4$ is not the right answer. Just put $x = 1$ and equate to $0$ to find $t$. — Rob Arthan, Mar 03 '18 at 23:33
The update means the second root is at $x=1$, so the book is saying $x^2-tx+9=(t-1)^2$ somehow? — , Mar 03 '18 at 23:48
@Goendo Please remember that you can choose an aswer among the given is the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work — user, Mar 09 '18 at 22:58

Benjamin Dickman · Answer 1 · 2018-03-03T23:38:20.010

5

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., $t=10$.

edited Mar 03 '18 at 23:38

answered Mar 03 '18 at 23:33

Benjamin Dickman

14,451

score 2 · Answer 2 · 2018-03-04T00:23:09.647

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 $x=1$.

edited Mar 04 '18 at 00:23

answered Mar 03 '18 at 23:29

How are you solving $(t + \sqrt{t^2 - 36})/2 = 1$ for $t$? – Rob Arthan Mar 03 '18 at 23:42
It should be self-explanatory for the person who asked the question. You get:$$\sqrt{t^2-36}=2-t$$ after which you square both sides. – Mar 03 '18 at 23:44
2

Way over the top! I'd be amazed if the OP found that "self-explanatory". There is a much simpler way - see the comments and other answers. – Rob Arthan Mar 03 '18 at 23:53
3

@idk $,\sqrt{t^2-36}=2-t,$ does not have $t=10$ as a root. – dxiv Mar 04 '18 at 00:19
@dxiv Let me put that as an extraneous. – Mar 04 '18 at 00:22
1

It's not extraneous. It's just wrong. – Rob Arthan Mar 04 '18 at 01:03

user · Answer 3 · 2018-03-03T23:34:48.043

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 $t=a+b=10$.

edited Mar 03 '18 at 23:34

answered Mar 03 '18 at 23:26

user

154,566

also in the question, $a$ was given as $\frac{1}{\sqrt a} + \sqrt a = 2$ By the way, $a = 1$ right? – Goendo Mar 03 '18 at 23:34
Yes, to check multiply by $\sqrt a$ and take $a=t^2$ we obtain $t^2-2t+1=(t-1)^2=0\implies t=1 \implies a=1$ – user Mar 03 '18 at 23:38

score 0 · Answer 4 · answered Mar 04 '18 at 03:54

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 $x=9$ $(t=10)$.

What is the value of $t$?

4 Answers4