# if-x-2-2cx-5d-0-has-root-a-and-b-also-x-2-2ax-5b-0-has-root-c-and-d-then-a-b-c-and-d-are-different-real-number-What-the-value-of-a-b-c-d-

Question Number 2859 by Syaka last updated on 29/Nov/15
$${if}\:{x}^{\mathrm{2}} \:−\:\mathrm{2}{cx}\:−\:\mathrm{5}{d}\:=\:\mathrm{0}\:{has}\:{root}\:{a}\:{and}\:{b}, \\$$$${also}\:{x}^{\mathrm{2}} \:−\:\mathrm{2}{ax}\:−\:\mathrm{5}{b}\:=\:\mathrm{0}\:{has}\:{root}\:{c}\:{and}\:{d} \\$$$${then}\:{a},\:{b},\:{c}\:{and}\:{d}\:{are}\:{different}\:{real}\:{number}. \\$$$${What}\:{the}\:{value}\:{of}\:{a}\:+\:{b}\:+\:{c}\:+\:{d}\:=\:? \\$$
Commented by Syaka last updated on 29/Nov/15
$${Nice}\:{One}\:{Sir}….\:{Thanks}\:{to}\:{Sir}\:{Rasheed}\:{and}\:{Sir}\:{Prakash}\:{for}\:{Solution} \\$$
Answered by Rasheed Soomro last updated on 29/Nov/15
$${x}^{\mathrm{2}} \:−\:\mathrm{2}{cx}\:−\:\mathrm{5}{d}\:=\:\mathrm{0}\:\:{a},{b}\:{are}\:{roots} \\$$$$\:{x}^{\mathrm{2}} \:−\:\mathrm{2}{ax}\:−\:\mathrm{5}{b}\:=\:\mathrm{0}\:\:\:{c},{d}\:{are}\:{roos} \\$$$${a}+{b}+{c}+{d}=? \\$$$${If}\:\:\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{has}\:\:\alpha\:\:{and}\:\:\beta\:\:\:{roots} \\$$$$\alpha+\beta=\frac{−{b}}{{a}}\:\:{and}\:\alpha\beta=\frac{{c}}{{a}} \\$$$${Sum} \\$$$${a}+{b}=\frac{−\left(−\mathrm{2}{c}\right)}{\mathrm{1}}=\mathrm{2}{c}……….\left({i}\right) \\$$$${c}+{d}=\frac{−\left(−\mathrm{2}{a}\right)}{\mathrm{1}}=\mathrm{2}{a}……….\left({ii}\right. \\$$$${Adding}\:\left({i}\right)\:{and}\:\left({ii}\right) \\$$$${a}+{b}+{c}+{d}=\mathrm{2}{c}+\mathrm{2}{a}\:=\mathrm{2}\left({a}+{c}\right)…………………\left(\mathrm{1}\right) \\$$$${a}+{c}={b}+{d} \\$$$${Product} \\$$$${ab}=−\mathrm{5}{d}\:{and}\:{cd}=−\mathrm{5}{b}\Rightarrow{abcd}=\mathrm{25}{bd}\Rightarrow{ac}=\mathrm{25}\Rightarrow{c}=\frac{\mathrm{25}}{{a}}…….\left(\mathrm{2}\right) \\$$$${From}\:\:\left(\mathrm{1}\right)\:\:{and}\:\:\left(\:\mathrm{2}\right) \\$$$${a}+{b}+{c}+{d}=\mathrm{2}\left({a}+\frac{\mathrm{25}}{{a}}\right) \\$$$${a}=\frac{−\left(−\mathrm{2}{c}\right)+\sqrt{\left(−\mathrm{2}{c}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{5}{d}\right)}}{\mathrm{2}\left(\mathrm{1}\right)}=\frac{\mathrm{2}{c}+\mathrm{2}\sqrt{{c}^{\mathrm{2}} +\mathrm{5}{d}}}{\mathrm{2}} \\$$$$\:\:\:\:{a}\:={c}+\sqrt{{c}^{\mathrm{2}} +\mathrm{5}{d}} \\$$$$\:\:\:\:{b}\:={c}−\sqrt{{c}^{\mathrm{2}} +\mathrm{5}{d}}\:\:\:\Rightarrow\:{a}−{b}=\mathrm{2}\sqrt{{c}^{\mathrm{2}} +\mathrm{5}{d}}\:\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}=\mathrm{4}{c}^{\mathrm{2}} +\mathrm{20}{d} \\$$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{4}{c}^{\mathrm{2}} =\mathrm{2}{ab}+\mathrm{20}{d}……………………..\left({A}\right) \\$$$${c}=\frac{−\left(−\mathrm{2}{a}\right)+\sqrt{\left(−\mathrm{2}{a}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{5}{b}\right)}}{\mathrm{2}\left(\mathrm{1}\right)}=\frac{\mathrm{2}{a}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{5}{b}}}{\mathrm{2}} \\$$$${c}={a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{5}{b}} \\$$$${d}={a}−\sqrt{{a}^{\mathrm{2}} +\mathrm{5}{b}}\:\:\Rightarrow\:{c}−{d}=\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{5}{b}}\:\Rightarrow{c}^{\mathrm{2}} +{d}^{\mathrm{2}} −\mathrm{2}{cd}=\mathrm{4}{a}^{\mathrm{2}} +\mathrm{20}{b} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow−\mathrm{4}{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{2}{cd}+\mathrm{20}{b}…………………….\left({B}\right) \\$$$$\mathcal{F}{or}\:{complete}\:{solution}\:\left({independant}\:{of}\:{a},{b},{c}\:{and}\:{d}\:\:{i}−{e}\:\:{constant}\right) \\$$$${see}\:{the}\:{comment}\:{of}\:\mathcal{S}{ir}\:{prakash}\:{jain}\:{below} \\$$$$\\$$
Commented by prakash jain last updated on 29/Nov/15
$$\mathrm{I}\:\mathrm{think}\:\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\:\mathrm{should}\:\mathrm{evaluate}\:\mathrm{to}\:\mathrm{a}\:\mathrm{constant} \\$$$$\mathrm{value}\:\mathrm{since}\:\mathrm{you}\:\mathrm{have}\:\mathrm{4}\:\mathrm{variables}\:\mathrm{and}\:\mathrm{4}\:\mathrm{equation} \\$$$$\mathrm{that}\:\mathrm{you}\:\mathrm{derived}. \\$$
Commented by Syaka last updated on 29/Nov/15
$${and}\:{a}\:+\:{b}\:+\:{c}\:+\:{d}\:{I}\:{need}\:{exact}\:{value},\:{Sir} \\$$
Commented by RasheedAhmad last updated on 29/Nov/15
$$\mathcal{N}\boldsymbol{{ice}}\:\boldsymbol{\mathcal{S}{ir}}! \\$$
Commented by prakash jain last updated on 29/Nov/15
$$\mathrm{Four}\:\mathrm{equation}\:\mathrm{given}\:\mathrm{by}\:\mathrm{Rasheed} \\$$$${a}+{b}=\mathrm{2}{c} \\$$$${c}+{d}=\mathrm{2}{a} \\$$$${ab}=−\mathrm{5}{d} \\$$$${cd}=−\mathrm{5}{b} \\$$$${c}=\frac{\mathrm{25}}{{a}} \\$$$${a}+{b}=\mathrm{2}{c}\Rightarrow{b}=\frac{\mathrm{50}−{a}^{\mathrm{2}} }{{a}} \\$$$${d}=\frac{−\mathrm{5}{b}}{{c}}=\frac{−\mathrm{5}\left(\mathrm{50}−{a}^{\mathrm{2}} \right)}{{a}\frac{\mathrm{25}}{{a}}}=−\frac{\mathrm{50}−{a}^{\mathrm{2}} }{\mathrm{5}} \\$$$${c}+{d}=\mathrm{2}{a} \\$$$${a}={b}={c}={d}=\mathrm{0}\:\mathrm{is}\:\mathrm{trival}\:\mathrm{solution}\:\:…\left({A}\right) \\$$$$\frac{\mathrm{25}}{{a}}−\frac{\mathrm{50}−{a}^{\mathrm{2}} }{\mathrm{5}}=\mathrm{2}{a} \\$$$$\mathrm{125}−\mathrm{50}{a}+{a}^{\mathrm{3}} =\mathrm{10}{a}^{\mathrm{2}} \\$$$${a}^{\mathrm{3}} −\mathrm{10}{a}^{\mathrm{2}} −\mathrm{50}{a}+\mathrm{125}=\mathrm{0} \\$$$${a}^{\mathrm{3}} +\mathrm{5}{a}^{\mathrm{2}} −\mathrm{15}{a}^{\mathrm{2}} −\mathrm{75}{a}+\mathrm{25}{a}+\mathrm{125}=\mathrm{0} \\$$$$\left({a}+\mathrm{5}\right)\left({a}^{\mathrm{2}} −\mathrm{15}{a}+\mathrm{25}\right)=\mathrm{0} \\$$$${a}=−\mathrm{5}\:{or}\:{a}=\frac{\mathrm{15}\pm\sqrt{\mathrm{225}−\mathrm{100}}}{\mathrm{2}}=\frac{\mathrm{15}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\$$$${a}=−\mathrm{5},\:{c}=−\mathrm{5},\:{b}=−\mathrm{5},\:{c}=−\mathrm{5}\:\:\:\:\:\:\:\:….\left(\mathrm{B}\right) \\$$$${a}=\frac{\mathrm{15}−\sqrt{\mathrm{5}}}{\mathrm{2}},{c}=\frac{\mathrm{25}}{{a}}=\frac{\mathrm{50}}{\mathrm{15}−\sqrt{\mathrm{5}}}=\frac{\mathrm{50}\left(\mathrm{15}+\sqrt{\mathrm{5}}\right)}{\mathrm{100}}=\frac{\mathrm{15}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\$$$$\:\:\:\:\:{b}=\mathrm{2}{c}−{a}=\frac{\mathrm{15}+\mathrm{15}\sqrt{\mathrm{5}}}{\mathrm{2}},{d}=\frac{\mathrm{15}−\mathrm{15}\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\:\:\:…\left({C}\right) \\$$$${similar}\:{you}\:{can}\:{get}\:{the}\:{last}\:{solution}\:{for}\:{a}=\frac{\mathrm{15}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\$$$${a}=\frac{\mathrm{15}+\sqrt{\mathrm{5}}}{\mathrm{2}},{c}=\frac{\mathrm{15}−\sqrt{\mathrm{5}}}{\mathrm{2}},{b}=\frac{\mathrm{15}−\mathrm{15}\sqrt{\mathrm{5}}}{\mathrm{2}},\:{d}=\frac{\mathrm{15}+\mathrm{15}\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:..\left({D}\right) \\$$$${C}\:\mathrm{ans}\:{D}\:\mathrm{are}\:\mathrm{only}\:\mathrm{valid}\:\mathrm{solutions}\:\mathrm{for} \\$$$$\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\mathrm{are}\:\mathrm{different}. \\$$$$\mathrm{in}\:\mathrm{both}\:\mathrm{cases} \\$$$$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=\mathrm{30} \\$$