Solve by factoring: Am I doing this right?
- Thread starter Zanfib
- Start date
Yes. You have not solved the equations as you were asked to do. Nor have you factored the expressions as was intended.
\(\displaystyle ab = 0 \implies\ a = 0 \ne b\ or\ b = 0 \ne a\ or\ a = 0 = b.\) Zero product property.
If you have an equation in the form of px^2 + qx + r = 0 and you can factor the quadratic into the form that
px^2 + qx + r = (jx - k)(mx - n) = 0, then the solutions are x = k/j and x = n/m.
Example
\(\displaystyle 6x^2 - 42x + 72 = 0 \implies 6x^2 - 42x + 72 = (3x - 9)(2x - 8) = 0 \implies 3x - 9 = 0\ or\ 2x - 8 = 0 \implies\)
\(\displaystyle x = 3\ or\ x = 4.\)
Let's check.
\(\displaystyle 6(3^2) - 42(3) + 72 = 6 * 9 - 126 + 72 = 54 + 72 - 126 = 126 - 126 = 0.\) CHECKS
\(\displaystyle 6(4^2) - 42(4) + 72 = 6 * 16 - 168 + 72 = 96 + 72 - 168 = 168 - 168 = 0.\) CHECKS
Now give your two problems another try, and we can confirm your answers.
Look fine to me
\(\displaystyle 11^2 - 5(11) - 66 = 121 - 55 - 66 = 121 - 121 = 0.\)
\(\displaystyle (-6)^2 - 5(-6) - 66 = 36 + 30 - 66 = 0.\)
\(\displaystyle 3\left(-\dfrac{2}{3}\right)^2 + 14\left(- \dfrac{2}{3}\right) + 8 = \dfrac{3 * 4}{9} + \dfrac{14 * (-2)}{3} + 8 = \dfrac{4 - 28 + 24}{3} = \dfrac{0}{3} = 0.\)
\(\displaystyle 3(-4)^2 + 14(-4) + 8 = 3 * 16 - 56 + 8 = 48 + 8 - 56 = 0.\)
Well done.
My fault, Mark. I said we would confirm his answers. I should have said that we would be happy to help if he continued to have trouble.
Yes. I was pretty confident that my new answer was correct, but I didn't want to just vanish without telling anyone that I had solved it.
Wisdom of the Ancients and all that.
mmm4444bot
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No -- my bad. I deleted my first post because I somehow (at the time of posting) missed seeing Jeff's confirmation