Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 33322 by Rio Mike last updated on 15/Apr/18

a) px^2 + 3x + q=0 has roots [((α+β)/(αβ))]× αβ find p and q if x^2 − 7x + 4 = 0 with real and distinct roots has same roots.. b)if α and β are roots of x^2 + kx +2k+8=0. a) find k if one root is twice the other.

$$\left.{a}\right)\:{px}^{\mathrm{2}} +\:\mathrm{3}{x}\:+\:{q}=\mathrm{0}\:{has}\:{roots}\: \\ $$$$\:\:\left[\frac{\alpha+\beta}{\alpha\beta}\right]×\:\alpha\beta \\ $$$${find}\:{p}\:{and}\:{q}\:{if}\: \\ $$$$\:\:{x}^{\mathrm{2}} −\:\mathrm{7}{x}\:+\:\mathrm{4}\:=\:\mathrm{0}\:{with}\:{real}\:{and}\: \\ $$$${distinct}\:{roots}\:{has}\:{same}\:{roots}.. \\ $$$$\left.{b}\right){if}\:\alpha\:{and}\:\beta\:{are}\:{roots}\:{of}\: \\ $$$$\:{x}^{\mathrm{2}} +\:{kx}\:+\mathrm{2}{k}+\mathrm{8}=\mathrm{0}. \\ $$$$\left.{a}\right)\:{find}\:{k}\:{if}\:{one}\:{root}\:{is}\:{twice}\:{the}\:{other}. \\ $$

Commented by Rio Mike last updated on 15/Apr/18

yes sir take it these way i tried solving it x^2 +kx+2k+8=0 α+β= −k and αβ= 2k+8 let one root be α and the other 2α SOR_n = α+2α= −k ⇒ ((3α)/(3 ))= ((−k)/3) α= ((−k)/3) POR_n α(2α)= 2k+8 2α^2 − 2k −8=0 2(((−k)/3))^2 − 2k − 8=0 k^2 −9k − 36 =0 k=3 or k=12

$${yes}\:{sir}\:{take}\:{it}\:{these}\:{way}\:{i}\:{tried}\:{solving}\:{it} \\ $$$$\:\:\:{x}^{\mathrm{2}} +{kx}+\mathrm{2}{k}+\mathrm{8}=\mathrm{0} \\ $$$$\alpha+\beta=\:−{k}\:\:\:{and}\:\alpha\beta=\:\mathrm{2}{k}+\mathrm{8} \\ $$$${let}\:{one}\:{root}\:{be}\:\alpha\:{and}\:{the}\:{other}\:\mathrm{2}\alpha \\ $$$${SOR}_{{n}} =\:\alpha+\mathrm{2}\alpha=\:−{k}\:\Rightarrow\:\frac{\mathrm{3}\alpha}{\mathrm{3}\:}=\:\frac{−{k}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha=\:\frac{−{k}}{\mathrm{3}} \\ $$$${POR}_{{n}} \:\alpha\left(\mathrm{2}\alpha\right)=\:\mathrm{2}{k}+\mathrm{8}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\alpha^{\mathrm{2}} −\:\mathrm{2}{k}\:−\mathrm{8}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2}\left(\frac{−{k}}{\mathrm{3}}\right)^{\mathrm{2}} −\:\mathrm{2}{k}\:−\:\mathrm{8}=\mathrm{0} \\ $$$$\:\:\:{k}^{\mathrm{2}} −\mathrm{9}{k}\:−\:\mathrm{36}\:=\mathrm{0} \\ $$$$\:\:\:\:\:{k}=\mathrm{3}\:{or}\:{k}=\mathrm{12} \\ $$

Commented by MJS last updated on 15/Apr/18

x^2 −7x+4=0 x=(7/2)±(√(((49)/4)−4))=(7/2)±((√(33))/2) α=((7−(√(33)))/2) β=((7+(√(33)))/2) px^2 +3x+q=0 x^2 +(3/p)x+(q/p)=0 (3/p)=−7 ⇒ p=−(3/7) (q/p)=4 ⇒ q=−((12)/7) −(3/7)x^2 +3x−((12)/7)=0

$${x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{4}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{7}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{49}}{\mathrm{4}}−\mathrm{4}}=\frac{\mathrm{7}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{33}}}{\mathrm{2}} \\ $$$$\alpha=\frac{\mathrm{7}−\sqrt{\mathrm{33}}}{\mathrm{2}} \\ $$$$\beta=\frac{\mathrm{7}+\sqrt{\mathrm{33}}}{\mathrm{2}} \\ $$$${px}^{\mathrm{2}} +\mathrm{3}{x}+{q}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{3}}{{p}}{x}+\frac{{q}}{{p}}=\mathrm{0} \\ $$$$\frac{\mathrm{3}}{{p}}=−\mathrm{7}\:\Rightarrow\:{p}=−\frac{\mathrm{3}}{\mathrm{7}} \\ $$$$\frac{{q}}{{p}}=\mathrm{4}\:\Rightarrow\:{q}=−\frac{\mathrm{12}}{\mathrm{7}} \\ $$$$−\frac{\mathrm{3}}{\mathrm{7}}{x}^{\mathrm{2}} +\mathrm{3}{x}−\frac{\mathrm{12}}{\mathrm{7}}=\mathrm{0} \\ $$

Commented by MJS last updated on 15/Apr/18

sorry somehow I thought it was x^3 ... I might need new glasses ;−) anyway your answer is right

$$\mathrm{sorry}\:\mathrm{somehow}\:\mathrm{I}\:\mathrm{thought}\:\mathrm{it}\:\mathrm{was} \\ $$$$\left.{x}^{\mathrm{3}} ...\:\mathrm{I}\:\mathrm{might}\:\mathrm{need}\:\mathrm{new}\:\mathrm{glasses}\:;−\right) \\ $$$$\mathrm{anyway}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{right} \\ $$

Commented by Rasheed.Sindhi last updated on 16/Apr/18

What is the use of [((α+β)/(αβ))]× αβ ?

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{use}\:\mathrm{of}\:\left[\frac{\alpha+\beta}{\alpha\beta}\right]×\:\alpha\beta\:? \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com