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 140974 by EnterUsername last updated on 14/May/21

$$\mathrm{If}\mathrm{the}\mathrm{equations}{\mathrm{x}}^2-3\mathrm{x}+a=0\mathrm{and}{\mathrm{x}}^2+a\mathrm{x}-3=0$$$$\mathrm{have}\mathrm{a}\mathrm{common}\mathrm{root},\mathrm{then}\mathrm{a}\mathrm{possible}\mathrm{value}\mathrm{of}a\mathrm{is}$$$$\left(\mathrm{A}\right)3\left(\mathrm{B}\right)1\left(\mathrm{C}\right)-2\left(\mathrm{D}\right)2$$

Commented by mohammad17 last updated on 14/May/21

$$$$$$x^2+ax-3=0\to x+\mathrm{a}-\frac{3}{\mathrm{x}}=0\to \left(1\right)$$$$$$$$x^2-3x+a=0\to \left(2\right)$$$$$$$$from\left(2\right)-\left(1\right)$$$$$$$$x^2-4x+\frac{3}{x}=0\Rightarrow x^3-4x^2+3=0$$$$$$$$x=1,x=\frac{3-\sqrt{21}}{2},x=\frac{3+\sqrt{21}}{2}$$$$$$$$since:x=1\Rightarrow a=2$$$$$$$$since:x=\frac{3+\sqrt{21}}{2}\Rightarrow a=-3cancel$$$$$$$$sincex=\frac{3-\sqrt{21}}{2}\Rightarrow a=-3cance$$$$$$$$\Rightarrow a=2$$

Commented by EnterUsername last updated on 14/May/21

$$\mathrm{Thanks}\mathrm{you}!$$

Commented by mohammad17 last updated on 14/May/21

$$youarewelcome$$

Answered by henderson last updated on 14/May/21

$$\mathrm{the}\mathrm{answer}\mathrm{is}a=2.$$

Answered by physicstutes last updated on 14/May/21

$$\mathrm{I}\mathrm{will}\mathrm{say}\mathrm{from}\mathrm{experience}a=2,\mathrm{but}\mathrm{if}\mathrm{you}\mathrm{need}\mathrm{a}\mathrm{proper}$$$$\mathrm{working}\mathrm{step},\mathrm{you}\mathrm{could}\mathrm{say}$$$$\mathrm{let}\mathrm{the}\mathrm{first}\mathrm{equation}\mathrm{have}\mathrm{roots}\alpha \mathrm{and}\beta$$$$\mathrm{and}\mathrm{let}\mathrm{the}\mathrm{second}\mathrm{equation}\mathrm{have}\mathrm{root}\beta \mathrm{and}\vartheta$$$$\mathrm{notice}\mathrm{that}\beta \mathrm{is}\mathrm{the}\mathrm{common}\mathrm{root}.$$$$\alpha +\beta =3.....\left(i\right)\mathrm{and}\alpha \beta =a.....\left(ii\right)$$$$\beta +\vartheta =-a.....\left(ii\right)\mathrm{and}\beta \vartheta =-3.....\left(iv\right)$$$$\mathrm{from}\left(iv\right)\vartheta =-\frac{3}{\beta }$$$$\Rightarrow \beta -\frac{3}{\beta }=-a$$$$\mathrm{also}\alpha =\frac{a}{\beta }$$$$\Rightarrow \frac{a}{\beta }+\beta =3\mathrm{and}\frac{3}{\beta }-\beta =a$$$$a+{\beta }^2=3\beta$$$$\mathrm{and}3-{\beta }^2=a\beta$$$$\mathrm{add}\mathrm{them}\mathrm{and}\mathrm{get}:3+a=\beta (3+a)\Rightarrow \beta =1$$$$\mathrm{now}1-\frac{3}{1}=-a\Rightarrow a=2$$

Commented by EnterUsername last updated on 14/May/21

$$\mathrm{Thanks}\mathrm{for}\mathrm{the}\mathrm{idea}!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com