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Which-of-the-following-expressions-are-positive-for-all-real-values-of-x-a-x-2-2x-5-b-x-2-2x-1-c-x-2-4x-2-d-2x-2-6x-5-




Question Number 37940 by Rio Mike last updated on 19/Jun/18
Which of the following   expressions are positive for  all real values of  x?  a) x^2 − 2x + 5   b) x^2 −2x−1   c) x^2 +4x+2      d) 2x^2 −6x + 5
$${Which}\:{of}\:{the}\:{following}\: \\ $$$${expressions}\:{are}\:{positive}\:{for} \\ $$$${all}\:{real}\:{values}\:{of}\:\:{x}? \\ $$$$\left.{a}\left.\right)\:{x}^{\mathrm{2}} −\:\mathrm{2}{x}\:+\:\mathrm{5}\:\:\:{b}\right)\:{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}\: \\ $$$$\left.{c}\left.\right)\:{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}\:\:\:\:\:\:{d}\right)\:\mathrm{2}{x}^{\mathrm{2}} −\mathrm{6}{x}\:+\:\mathrm{5} \\ $$
Answered by MrW3 last updated on 19/Jun/18
a) x^2 −2x+5=x^2 −2x+1+4=(x−1)^2 +4≥4>0  b) x^2 −2x−1=x^2 −2x+1−2=(x−1)^2 −3≥−3  c) x^2 +4x+2=x^2 +4x+4−2=(x+2)^2 −2≥−2  d) 2x^2 −6x+5=2(x^2 −3x+(9/4))+(1/2)=2(x−(3/2))^2 +(1/2)≥(1/2)>0  ⇒answer is a) and d)
$$\left.{a}\right)\:{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{5}={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}+\mathrm{4}=\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}\geqslant\mathrm{4}>\mathrm{0} \\ $$$$\left.{b}\right)\:{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}−\mathrm{2}=\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}\geqslant−\mathrm{3} \\ $$$$\left.{c}\right)\:{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}={x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}−\mathrm{2}=\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}\geqslant−\mathrm{2} \\ $$$$\left.{d}\right)\:\mathrm{2}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{5}=\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\frac{\mathrm{9}}{\mathrm{4}}\right)+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{2}\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\geqslant\frac{\mathrm{1}}{\mathrm{2}}>\mathrm{0} \\ $$$$\left.\Rightarrow\left.{answer}\:{is}\:{a}\right)\:{and}\:{d}\right) \\ $$

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