Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 20053 by Tinkutara last updated on 21/Aug/17

If (4a + c)^2  ≤ 4b^2  then one root of  ax^2  + bx + c = 0 lies in  (1) (−2, 2)  (2) (−1, 1)  (3) (−∞, −2)  (4) (2, ∞)

$$\mathrm{If}\:\left(\mathrm{4}{a}\:+\:{c}\right)^{\mathrm{2}} \:\leqslant\:\mathrm{4}{b}^{\mathrm{2}} \:\mathrm{then}\:\mathrm{one}\:\mathrm{root}\:\mathrm{of} \\ $$$${ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{lies}\:\mathrm{in} \\ $$$$\left(\mathrm{1}\right)\:\left(−\mathrm{2},\:\mathrm{2}\right) \\ $$$$\left(\mathrm{2}\right)\:\left(−\mathrm{1},\:\mathrm{1}\right) \\ $$$$\left(\mathrm{3}\right)\:\left(−\infty,\:−\mathrm{2}\right) \\ $$$$\left(\mathrm{4}\right)\:\left(\mathrm{2},\:\infty\right) \\ $$

Answered by ajfour last updated on 21/Aug/17

(4a+c)^2 −4b^2  ≤ 0  ⇒  (4a+2b+c)(4a−2b+c) ≤0  ⇒  f(2)f(−2)≤0  ⇒  f(x) changes sign or is zero   in [−2,2], hence one root  lies in [−2, 2].

$$\left(\mathrm{4}{a}+{c}\right)^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} \:\leqslant\:\mathrm{0} \\ $$$$\Rightarrow\:\:\left(\mathrm{4}{a}+\mathrm{2}{b}+{c}\right)\left(\mathrm{4}{a}−\mathrm{2}{b}+{c}\right)\:\leqslant\mathrm{0} \\ $$$$\Rightarrow\:\:{f}\left(\mathrm{2}\right){f}\left(−\mathrm{2}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow\:\:{f}\left({x}\right)\:{changes}\:{sign}\:{or}\:{is}\:{zero}\: \\ $$$${in}\:\left[−\mathrm{2},\mathrm{2}\right],\:{hence}\:{one}\:{root} \\ $$$${lies}\:{in}\:\left[−\mathrm{2},\:\mathrm{2}\right]. \\ $$

Commented by Tinkutara last updated on 21/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com