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 81771 by Power last updated on 15/Feb/20

Commented by Power last updated on 15/Feb/20

solution

$$\mathrm{solution} \\ $$

Commented by Power last updated on 15/Feb/20

What value of the parameter  ↑  will have 2 roots ?

$$\mathrm{What}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parameter}\:\:\uparrow\:\:\mathrm{will}\:\mathrm{have}\:\mathrm{2}\:\mathrm{roots}\:? \\ $$

Commented by mr W last updated on 15/Feb/20

two roots when −4≤a<2  one root when a=2  no root when a<−4 or a>2

$${two}\:{roots}\:{when}\:−\mathrm{4}\leqslant{a}<\mathrm{2} \\ $$$${one}\:{root}\:{when}\:{a}=\mathrm{2} \\ $$$${no}\:{root}\:{when}\:{a}<−\mathrm{4}\:{or}\:{a}>\mathrm{2} \\ $$

Answered by mr W last updated on 15/Feb/20

let f(x)=(√(4−∣2x∣))  let g(x)=x^2 +a  both functions are even, i.e.   f(−x)=f(x), g(−x)=g(x)  we only need to look at x≥0.  f(x)=(√(4−2x)) :  0≤x≤2  f(0)=2=maximum  f(2)=0=minimum  g(x)=x^2 +a :  this is a parabola opening upwards,  its lowest point is (0,a).    such that the eqn. has two roots, the  curve g(x) must intersect the curve g(x)  at two points.  we have following cases:    case 1: a>2, g(x) is too high and can′t  intersect the curve f(x) ⇒no root.    case 2: a=2, the lowest point of g(x)   coincides with the topmost point of  f(x), one intersection point ⇒one  root.    case 3: −4≤a<2, g(x) intersects with  f(x) at two points ⇒two roots.    case 4: a=−4, y=x^2 −4, y=0 ⇒x=±2.  f(x) intersects with g(x) at (−2,0) and  (2,0)⇒two roots.    case 5: a<−4, g(x) is too low and can′t  intersect with f(x)⇒no root.

$${let}\:{f}\left({x}\right)=\sqrt{\mathrm{4}−\mid\mathrm{2}{x}\mid} \\ $$$${let}\:{g}\left({x}\right)={x}^{\mathrm{2}} +{a} \\ $$$${both}\:{functions}\:{are}\:{even},\:{i}.{e}.\: \\ $$$${f}\left(−{x}\right)={f}\left({x}\right),\:{g}\left(−{x}\right)={g}\left({x}\right) \\ $$$${we}\:{only}\:{need}\:{to}\:{look}\:{at}\:{x}\geqslant\mathrm{0}. \\ $$$${f}\left({x}\right)=\sqrt{\mathrm{4}−\mathrm{2}{x}}\:: \\ $$$$\mathrm{0}\leqslant{x}\leqslant\mathrm{2} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{2}={maximum} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{0}={minimum} \\ $$$${g}\left({x}\right)={x}^{\mathrm{2}} +{a}\:: \\ $$$${this}\:{is}\:{a}\:{parabola}\:{opening}\:{upwards}, \\ $$$${its}\:{lowest}\:{point}\:{is}\:\left(\mathrm{0},{a}\right). \\ $$$$ \\ $$$${such}\:{that}\:{the}\:{eqn}.\:{has}\:{two}\:{roots},\:{the} \\ $$$${curve}\:{g}\left({x}\right)\:{must}\:{intersect}\:{the}\:{curve}\:{g}\left({x}\right) \\ $$$${at}\:{two}\:{points}. \\ $$$${we}\:{have}\:{following}\:{cases}: \\ $$$$ \\ $$$${case}\:\mathrm{1}:\:{a}>\mathrm{2},\:{g}\left({x}\right)\:{is}\:{too}\:{high}\:{and}\:{can}'{t} \\ $$$${intersect}\:{the}\:{curve}\:{f}\left({x}\right)\:\Rightarrow{no}\:{root}. \\ $$$$ \\ $$$${case}\:\mathrm{2}:\:{a}=\mathrm{2},\:{the}\:{lowest}\:{point}\:{of}\:{g}\left({x}\right)\: \\ $$$${coincides}\:{with}\:{the}\:{topmost}\:{point}\:{of} \\ $$$${f}\left({x}\right),\:{one}\:{intersection}\:{point}\:\Rightarrow{one} \\ $$$${root}. \\ $$$$ \\ $$$${case}\:\mathrm{3}:\:−\mathrm{4}\leqslant{a}<\mathrm{2},\:{g}\left({x}\right)\:{intersects}\:{with} \\ $$$${f}\left({x}\right)\:{at}\:{two}\:{points}\:\Rightarrow{two}\:{roots}. \\ $$$$ \\ $$$${case}\:\mathrm{4}:\:{a}=−\mathrm{4},\:{y}={x}^{\mathrm{2}} −\mathrm{4},\:{y}=\mathrm{0}\:\Rightarrow{x}=\pm\mathrm{2}. \\ $$$${f}\left({x}\right)\:{intersects}\:{with}\:{g}\left({x}\right)\:{at}\:\left(−\mathrm{2},\mathrm{0}\right)\:{and} \\ $$$$\left(\mathrm{2},\mathrm{0}\right)\Rightarrow{two}\:{roots}. \\ $$$$ \\ $$$${case}\:\mathrm{5}:\:{a}<−\mathrm{4},\:{g}\left({x}\right)\:{is}\:{too}\:{low}\:{and}\:{can}'{t} \\ $$$${intersect}\:{with}\:{f}\left({x}\right)\Rightarrow{no}\:{root}. \\ $$

Commented by mr W last updated on 15/Feb/20

Commented by Power last updated on 15/Feb/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 15/Feb/20

is this a valid proof according to you?

$${is}\:{this}\:{a}\:{valid}\:{proof}\:{according}\:{to}\:{you}? \\ $$

Commented by Power last updated on 15/Feb/20

yes sir

$$\mathrm{yes}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com