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Homework Statement
Question + attempt:
Homework Equations
The Attempt at a Solution
Why is it that when I expand the factored form, I don't get the original equation?
You get m^{2} + (3/2)m + 1 = 0, which is equivalent to the original equation. Just multiply both sides by 2.Homework Statement
Question + attempt:
Homework Equations
The Attempt at a Solution
Why is it that when I expand the factored form, I don't get the original equation?
Let's say we have $$2{ m }^{ 2 }+3m+1=60$$.You get m^{2} + (3/2)m + 1 = 0, which is equivalent to the original equation. Just multiply both sides by 2.
No. The Quadratic Formula requires that your equation be in the form ax^{2} + bx + c = 0, with a ≠ 0.Let's say we have $$2{ m }^{ 2 }+3m+1=60$$.
Would I be able to apply the quadratic formula to the left hand-side of the equation only?
What do you think? These look like textbook problems. Our policy here is that we don't do the work for students. Tell me what you think and why, and I'll tell you if you're right or where you went wrong.How true is the below?
They're not problems. They're statements that I wrote.What do you think? These look like textbook problems. Our policy here is that we don't do the work for students. Tell me what you think and why, and I'll tell you if you're right or where you went wrong.
The general form of a quadratic function is y = ax^{2} + bx + c. The right side of this can be written as a(x -r)(x -s). It's possible that r and s are complex numbers, though.They're not problems. They're statements that I wrote.
If I recall correctly, in the 10th grade, the teacher said the general form of a quadratic equation is ##y=(x-r)(x-s)## and in the 11th grade, the teacher said the general form of a cubic equation is ##y=(x-r)(x-s)(x-t)##.
The general form of a cubic function is y = ax^{3} + bx^{2} + cx + d. The right side can be written in factored form as a(x - r)(x - s)(x - t), with possibly some of the roots being complex. Complex roots come in pairs, so we can't have more than two complex roots in a cubic equation.However, in the 11th grade, the teacher changed it to ##y=a(x-r)(x-s)(x-t)## later.
So what is the point of the a coefficient if you can always divide it out?Note that 2(m + 1)(m + 1/2) = 0 and (m + 1)(m + 1/2) = 0 are equivalent equations. They both have exactly the same solutions.