Question 115339
This is in response to a follow up from the student to clarify.  My original solution remains below:

m actually equals two numbers, 1 and - 6.

The quadratic equation deals with the coefficients of the equation you are trying to solve.  Your equation is some coefficient a * m^2 + some coefficient b * m - 6, which we will call c.  Remember that a coefficient is the number that falls before your x term.  We have two x terms, x^2 and x, so two coefficients, a and b.

In general form, the quadratic equation states that m = ( b + or - the square root of (b^2 - 4ac)) all divided by 2*a.

So the quadratic equation first tells you to first solve for b^2 - 4 * a * c.  For your specific equation: b, the number that appears before "m", is 5.  a, the number that appears before m^2, is 1.  And finally c, is -6.  Therefore b^2 = 5^2 = 25, and 4 * a * c = 4 * 1 * -6 = -24.  So b^2 - 4*a*c = 25 - (-24) = 25 + 24 = 49.

Next, we need to take the square root of that.  The square root of 49 is 7.

After this, we need to split into two different equations: -b + 7 and -b - 7.  Knowing that b is 5, -b must be -5, so we have -5 + 7 =2 and -5 - 7 = -12.

Finally, we need to divide both equations by 2 * a, or 2*1 = 2.  2/2 = 1, so your first solution is 1.  -12/2 = -6, so your second solution is -6.  Your answer is m = 1 and m = -6.

Let's plug in to make sure:  1(1)^2 + 5(1) - 6 = 1 + 5 - 6 = 6- 6 = 0.  This fits your original equation so 1 is definitely a solution.
1(-6)^2 + 5(-6) - 6 = 36 - 30 - 6 = 6 - 6 = 0, so -6 is also definitely a solution.

Therefore your answers are m = 1 and m = -6.

It may be a little confusing that an equation yields two solutions.  However, what you are doing is solving for when the given parabola crosses the line x= 0.  Since a parabola is symmetrical, this happens two times, yielding two solutions.


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Quadratic equations are typically expressed in terms of x, however the variable you use does not matter (in this case m).

*[invoke quadratic "m", 1, 5, -6]