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# Linear and quadratic systems — Harder example

Watch Sal work through a harder Linear and quadratic systems problem.

## Want to join the conversation?

• In a quadratic equation of form
ax^2+bx+c=0
the product of their roots is c/a.
So you dont need to solve them to find the product of two roots
• We could just use Vieta Theorem to find the product of the roots. It's quite more convenient.
• Another one is that the sum of the solutions of a quadratic in that form is -b/a.
• The SAT answer grid does not have an option for a negative answer. why is the answer here negative?
• An answer on the SAT won't be negative if you won't be able to answer it. :)
• you could have solved the question without even factorizing/solving the equation as the coefficient of X^0 that is -2 is already the product of the two X co-ordinates.
• What do you mean by the `coefficient of X^0`? The constant term in the equation?
• math is so delicious i love it
• Couldn't we just use the splitting method instead
• how do we know when to use the quadratic formula?
like why didnt we just do 2x^2+11x-6 =0
a.b = -12
a+b = 11
• A good rule of thumb is to spend maybe 30 seconds max on trying to think of factors. If they're not coming to you, go ahead and use the quadratic formula. The quadratic formula always works, but factoring can sometimes be faster. This means that both approaches will get you a correct answer, but how often you factor vs use quadratic formula depends on how good you are at factoring.
Here, we could factor the equation like you said, by breaking up the middle term into a +12x and a -x.
• How do you know when to substitute and when to set the equations equal to each other? For a simple problem like this where both equations equal Y I understand, but what if they don't?
• Pro tip: just use desmos graphing calculator and check the coordinates where the two equations intersect.