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Prove that the root of x^2+(1-k)x+k-3=0 are real for all real values of k |
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Answer» ROOTS of a quadratic EQUATION are real WHENEVER the discriminant is non-negative. For the function given, the discriminant is (1+k)2–4(1)(−k) , which SIMPLIFIES as k2+6k+1 . To determine when this is non-negative, we note that its zeros are at −6±36–4√2 , which are −3–22–√ and −3+22–√ . Since this expression is quadratic, it can be graphed as a parabola which opens upward, and so is non-negative to the left of the first ZERO, and to the right of the second zero. Consequently, the roots of the original function are real for values of k in (−∞,−3–22–√]∪[−3+22–√,∞) , so NOT for all values of k. |
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