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 147978 by mathdanisur last updated on 24/Jul/21

3x^2 - 7 ∣x∣ + m - 5 = 0 At what value of m can the equation have three roots

$$\mathrm{3}{x}^{\mathrm{2}} \:-\:\mathrm{7}\:\mid{x}\mid\:+\:{m}\:-\:\mathrm{5}\:=\:\mathrm{0}\: \\ $$ $${At}\:{what}\:{value}\:{of}\:\boldsymbol{{m}}\:{can}\:{the}\:{equation} \\ $$ $${have}\:{three}\:{roots} \\ $$

Answered by Olaf_Thorendsen last updated on 24/Jul/21

In R, the equation can have two opposite solutions for x<0 (∣x∣ = −x) and two opposite solutions for x>0 (∣x∣ = x). To obtain exactly three roots, one root needs to be 0 (to be equal to its opposite). That occurs when (c/a) = ((m−5)/3) = 0 ⇒ m = 5. In this case 3x^2 = 7∣x∣ ⇒ x = 0, x = ±(7/3)

$$\mathrm{In}\:\mathbb{R},\:\mathrm{the}\:\mathrm{equation}\:\mathrm{can}\:\mathrm{have}\:\mathrm{two} \\ $$ $$\mathrm{opposite}\:\mathrm{solutions}\:\mathrm{for}\:{x}<\mathrm{0}\:\left(\mid{x}\mid\:=\:−{x}\right) \\ $$ $$\mathrm{and}\:\mathrm{two}\:\mathrm{opposite}\:\mathrm{solutions}\:\mathrm{for}\:{x}>\mathrm{0} \\ $$ $$\left(\mid{x}\mid\:=\:{x}\right). \\ $$ $$ \\ $$ $$\mathrm{To}\:\mathrm{obtain}\:\mathrm{exactly}\:\mathrm{three}\:\mathrm{roots},\:\mathrm{one}\:\mathrm{root} \\ $$ $$\mathrm{needs}\:\mathrm{to}\:\mathrm{be}\:\mathrm{0}\:\left(\mathrm{to}\:\mathrm{be}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{its}\right. \\ $$ $$\left.\mathrm{opposite}\right). \\ $$ $$\mathrm{That}\:\mathrm{occurs}\:\mathrm{when}\:\frac{{c}}{{a}}\:=\:\frac{{m}−\mathrm{5}}{\mathrm{3}}\:=\:\mathrm{0} \\ $$ $$\Rightarrow\:{m}\:=\:\mathrm{5}. \\ $$ $$\mathrm{In}\:\mathrm{this}\:\mathrm{case}\:\mathrm{3}{x}^{\mathrm{2}} \:=\:\mathrm{7}\mid{x}\mid \\ $$ $$\Rightarrow\:{x}\:=\:\mathrm{0},\:{x}\:=\:\pm\frac{\mathrm{7}}{\mathrm{3}} \\ $$

Commented bymathdanisur last updated on 25/Jul/21

Thankyou Ser

$${Thankyou}\:{Ser} \\ $$

Answered by Rasheed.Sindhi last updated on 25/Jul/21

3x^2 - 7 ∣x∣ + m - 5 = 0 ⇒ { ((i:3x^2 - 7x =− m + 5)),((ii:3x^2 + 7x= − m + 5)) :}⇒3x^2 ^(×) −7x=3x^2 ^(×) +7x ⇒x=0 (common root of i & ii) ∵ i & ii have one root common ∴ They have total three roots ∴ The original equation has 3 roots 3x^2 - 7 ∣x∣ + m - 5 = 0 ⇒3(0)^2 −7∣0∣+m−5=0 ⇒m−5=0⇒m=5

$$\mathrm{3}{x}^{\mathrm{2}} \:-\:\mathrm{7}\:\mid{x}\mid\:+\:{m}\:-\:\mathrm{5}\:=\:\mathrm{0}\: \\ $$ $$\Rightarrow\begin{cases}{{i}:\mathrm{3}{x}^{\mathrm{2}} \:-\:\mathrm{7}{x}\:=−\:{m}\:+\:\mathrm{5}}\\{{ii}:\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{7}{x}=\:−\:{m}\:+\:\mathrm{5}}\end{cases}\Rightarrow\overset{×} {\mathrm{3}{x}^{\mathrm{2}} }−\mathrm{7}{x}=\overset{×} {\mathrm{3}{x}^{\mathrm{2}} }+\mathrm{7}{x} \\ $$ $$\Rightarrow{x}=\mathrm{0}\:\left({common}\:{root}\:{of}\:{i}\:\&\:{ii}\right) \\ $$ $$\because\:\:{i}\:\:\:\&\:\:\:{ii}\:\:\:\:{have}\:{one}\:{root}\:\:{common} \\ $$ $$\therefore\:\mathcal{T}{hey}\:{have}\:{total}\:{three}\:{roots} \\ $$ $$\therefore\:\mathcal{T}{he}\:{original}\:{equation}\:{has}\:\mathrm{3}\:{roots} \\ $$ $$\mathrm{3}{x}^{\mathrm{2}} \:-\:\mathrm{7}\:\mid{x}\mid\:+\:{m}\:-\:\mathrm{5}\:=\:\mathrm{0}\: \\ $$ $$\Rightarrow\mathrm{3}\left(\mathrm{0}\right)^{\mathrm{2}} −\mathrm{7}\mid\mathrm{0}\mid+{m}−\mathrm{5}=\mathrm{0} \\ $$ $$\Rightarrow{m}−\mathrm{5}=\mathrm{0}\Rightarrow{m}=\mathrm{5} \\ $$

Commented bymathdanisur last updated on 25/Jul/21

Thank you Ser

$${Thank}\:{you}\:{Ser} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com