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The Discriminant of a Quadratic Equation  Problem 3
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
The discriminant of a quadratic tells us the number and type of solutions to that equation and for this particular problem what we’re looking at is a problem with a variable k. We’re trying to find k so that we have one solution to this problem.
Discriminant is b² minus 4ac which is just what’s underneath the square root in your quadratic formula and for, in order for us to have one solution what we know is that we can’t have a square root or5 anything associated with that term at all, because with plus r minus that’s square root we need to plus or minus nothing in order so we just have the negative b over 2a left. The one solution tells me I need to have my discriminant equal to zero. If it’s not too imaginary I would know that my discriminant has to be less than zero, so I would say two rationals and wouldn’t need to set it equal to a perfect square. A number of different scenarios hold true but for this one we’re looking for one solution which tells me my discriminant is equal to zero.
From here we plug in our ab and our c and solve for k so b² is just going to be our k. This is going to give us, k² minus 4 a is 9 and c is 4 and this is going to equal to zero. Solving this out, k² minus 36 times 4, I believe is 144 but let’s just double check, 36 times 4, yeah, 144, 144 is equal to zero. From here we have options we could either add the 144 over take the square root, always remember your plus and minus or we could factor this out. Either one is perfectly fine, I am going to let’s say bring it over to the other side, so we have k² is equal to 144, square root both sides, when we use the square root as a tool, we need to include plus or minus giving us k is equal to plus or minus 12.
For this particular problem there’s actually two values we could have for k that would give this equation one solution.
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Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
Sample Problems (4)
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