sungkwo said:
hi, can you help me out with these questions
1. if k does not equal 0, solve the quadratic inequation for x(if x is all real)
kx^2+kx+2k<0
2. If k does not equal 1, solve the quadratic inequation for x(if x is all real)
2(x^2+1)>4x-a^2
Thanks
1) Did you type the question correctly? Seems rather unusual for 2 unit level.
kx² + kx + 2k < 0
k(x² + x + 2) < 0
But for x² + x + 2, Δ = 1 - 4(1)(2) = - 7 < 0
Hence, x² + x + 2 is positive definite which means x² + x + 2 > 0 for all real x.
Therefore, for k(x² + x + 2) to be negative k must be negative. Hence, we have no real solutions if k > 0 and all real solutions of x if k < 0...lol
2) 2(x² + 1) > 4x - a²
2x² + 2 > 4x - a²
2(x² - 2x + 2) > - a²
For x² - 2x + 2, Δ = 4 - 4(1)(2) = - 4 < 0
Hence, x² - 2x + 2 is positive definite which means x² - 2x + 2 > 0 for all real x.
Since a² > 0 (for non-zero a) for all real a, thus - a² < 0.
Since 2(x² - 2x + 2) > 0 always and positive numbers are always greater than negative numbers, then the inequality holds true for all real x.
