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# Algebra

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Let \$r\$ and \$s\$ be the roots of \$2x^2 + 5x - 13 = x^2 - 4x + 7.\$ Find \$r^2 + s^2.\$

Jun 4, 2024

#1
+759
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First, let's write a quadratic equation to find the two roots.

We have \(x^{2}+9x-20=0\)

Notice that \(r^2 + s^2 = (r+s)^2-2rs\)

Now, let's take a break from this. If we had roots r and s, we would have

\((x-r)(x-s)\\ x^2-sx-rx+rs\\ x^2-(r+s)x+rs\)

Using the equation from above, we find that, r+s is equal to -9 and rs is equal to -20.

Plugging that in to \( (r+s)^2-2rs\), we get\((-9)^2-2(-20) = 81+40 = 121\)

Thanks! :)\(\)

Jun 4, 2024

#1
+759
+1

First, let's write a quadratic equation to find the two roots.

We have \(x^{2}+9x-20=0\)

Notice that \(r^2 + s^2 = (r+s)^2-2rs\)

Now, let's take a break from this. If we had roots r and s, we would have

\((x-r)(x-s)\\ x^2-sx-rx+rs\\ x^2-(r+s)x+rs\)

Using the equation from above, we find that, r+s is equal to -9 and rs is equal to -20.

Plugging that in to \( (r+s)^2-2rs\), we get\((-9)^2-2(-20) = 81+40 = 121\)