I am in algebra II and I am unsure how to solve this! My teacher got -15.49 < b < 15.49 but I have no idea how. Please show me with steps on how to solve, Thanks so much!!
Answers (2)
You know the equation x1,x2 = (-b +- sqrt(b^2 - 4ac)) / 2a .
Now, when that square root part is 0, that means you have only one solution (the +- is negated). When it's >0, you have two, and <0 you have zero (ie both are complex).
So, in order to find the range of no real solutions, solve the inequality b^2 - 4ac < 0.
Dealing with 2nd degree polynomials, you should receive either v1 < x < v2 , or (x > v1 or x < v2); if you ever struggle with remembering which is which, test a couple values in either range and see if they satisfy the original condition. Checking the solution is always a good idea.
Negative b is the place to start,
Plus or minus will show you're smart,
The radical sits so gent-i-ly,
On b square minus 4ac
Now we're done ... except to say,
The whole thing sits on top of 2a
x = (-b ± √(b^2 - 4ac))/2a
"Not real" means (b^2 - 4ac) is negative. Solve that and you have your answer.
Thank you soooo much!!!!