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Question: Given that $r$ is a root of the quadratic equation $x^2−3x−5=0$, find the value of $2r^2−6r+1$

Thx a lot, because I can't see the pattern of this quadratic equation, so I am struggling a lot to use $r={3±√29\over 2}$ in $2r^2−6r+1$

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    $\begingroup$ Any thoughts? Hint, what's the value of $2r^2-6r-10$? $\endgroup$ – lulu Sep 13 '17 at 15:30
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    $\begingroup$ So what is your thinking, and what have you tried? What do you know about $r$ which might help you to evaluate the target expression? $\endgroup$ – Mark Bennet Sep 13 '17 at 15:30
  • $\begingroup$ Hint: If $r$ is a root of the equation then $r$ will satisfy the equation !! $\endgroup$ – Donald Splutterwit Sep 13 '17 at 15:33
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$$x^2-3x=5$$

$r$ is the value that makes that ^ equation correct. So:

$$r^2-3r=5$$

$$2r^2-6r=10$$

$$2r^2-6r+1=11$$

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One way is to literally find $r$, which isn't too hard since $x^2-3x-5=0$ has roots $x=\frac{3\pm\sqrt{29}}{2}$.

The way they probably want you to solve this question is to recognize that $r$ satisfies $r^2-3r-5=0$. Multiplying by $2$ yields $2r^2-6r-10=0$. Can you continue?

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Hint : $2x^2 - 6x = 2(x^2-3x)$.

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Even if you don't see the pattern, you can always use the relation $$r^2 - 3r - 5=0$$ to replace $r^2$ by $3r+5$ in the expression you want to evaluate.

In the worst case, since the expression to be evaluated is a quadratic polynomial in $r$, if $r$ doesn't get caneceled in the simplification, you'll end up with a linear expression instead of a quadratic one, so it's still a good strategy.

Thus, \begin{align*} r^2-3r - 5 &= 0\\[4pt] \implies\;r^2 &= 3r +5\\[4pt] \implies\;2r^2-6r+1&=2(3r+5)-6r+1\\[4pt] &=(6r+10)-6r +1\\[4pt] &=11\\[4pt] \end{align*}

But here's an observation . . .

Let's say at the outset, you were considering using the quadratic formula to solve for $r$. But the coefficients are real and the discriminant is $$b^2 - 4ac = (-3)^2-4(1)(-5) = 29$$ so even without solving, you know the equation has two distinct real roots.

But now, if you change your mind about using the quadratic formula, switching instead to the approach where you replace $r^2$ by $3r+5$, then even before making the substitution, it's clear that the result will simplify to either a constant or a polynomial of degree $1$ in $r$.

But if you assume that the answer is unique, then, since the problem didn't specify which root $r$, the expression can't end up being degree $1$ in $r$. That's because a polynomial of degree $1$ can't realize the same value twice, so must yield a different result for each root, contrary to the assumption of uniqueness.

The upshot of this observation is that you expect that in the process of simplification, the $r$ terms will cancel.

That said, you still have to do the work for the evaluation, but when the $r$ terms do cancel, then, based on this thought process, it shouldn't come as a surprise.

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