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Q:

Solve $x^2 – 7x +9 = 0$

$$x^2 – 7x +9 = 0$$

Factorize and replace $x$ by $t$
$$t^2 – 7t +9 = 0$$
$$(t-3)(t+2) = 0$$
So the solutions are $t=3$ or $t=-2$. Is this the correct way to find the solutions?

A:

It is pretty much the correct way. But you can take it a step further. There is only one real solution: $t = 3$.
Observe that when you factorize, you get the usual factors of $x – 3$ and $x + 2$. However, the number $x = 3$ was in the original polynomial as well. This means that you need only consider the factorization of the polynomial after you have factored out the linear factor $x – 3$. We can see that it is the same as the original factorization: $$\left(x – 3\right)^2 = x^2 – 6x + 9$$
So the only real solution for $x$ is $x = 3$. And for this to be a solution, it also has to be a solution for $t$, so we find that $t = 3$ as well.

A:

You are correct that all the solutions will be of the form $t = \pm 3$ and all are therefore real. You can see this by examining the coefficient of the quadratic term. In the original equation, we have:
$$x^2 – 7x + 9 = (x – 3)(x – 3) = 0$$
The quadratic term will be $x^2$, since $x – 3$ is a linear term and the constant term will be $9$, since the first factor is $x – 3$ and the second factor is $1$.

A:

$$x^2-7x+9=x(x-3)$$
$$(x-3)^2-7(x-3)+9=(x-3)(x-3-7)=0$$
So $$x=3 \,\,\, or \,\,\,x=-3$$
In general \$(x-3)(

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