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The-quadratic-equations-x-2-6x-a-0-and-x-2-cx-6-0-have-one-root-in-common-The-other-roots-of-the-first-and-second-equations-are-integers-in-the-ratio-4-3-Then-find-the-common-root-

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Question Number 20118 by Tinkutara last updated on 22/Aug/17
The quadratic equations x^2 − 6x + a = 0 and x^2 − cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then, find the common root.
$$\mathrm{The}\:\mathrm{quadratic}\:\mathrm{equations}\:{x}^{\mathrm{2}} \:−\:\mathrm{6}{x}\:+\:{a}\:=\:\mathrm{0} \\ $$$$\mathrm{and}\:{x}^{\mathrm{2}} \:−\:{cx}\:+\:\mathrm{6}\:=\:\mathrm{0}\:\mathrm{have}\:\mathrm{one}\:\mathrm{root}\:\mathrm{in} \\ $$$$\mathrm{common}.\:\mathrm{The}\:\mathrm{other}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{and}\:\mathrm{second}\:\mathrm{equations}\:\mathrm{are}\:\mathrm{integers}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{4}\::\:\mathrm{3}.\:\mathrm{Then},\:\mathrm{find}\:\mathrm{the}\:\mathrm{common} \\ $$$$\mathrm{root}. \\ $$
Answered by ajfour last updated on 22/Aug/17
Let the common root be α αβ=a , α+β=6 , αγ=6 , α+γ=c ⇒ β−γ = 6−c (β/γ) = (a/6)=(4/3) ⇒ a=8 α+β=6 ; αβ=8 , ⇒ α=2,4 A check: if α=2, then β=4 , and γ=(3/4)β=3 if α=4, then β=2, and γ=(3/4)β=(3/2) ; not an integer. Hence α=2 . c=α+γ =2+3=5 .
$${Let}\:{the}\:{common}\:{root}\:{be}\:\alpha \\ $$$$\alpha\beta={a}\:,\:\:\alpha+\beta=\mathrm{6}\:,\:\alpha\gamma=\mathrm{6}\:,\:\alpha+\gamma={c} \\ $$$$\Rightarrow\:\:\beta−\gamma\:=\:\mathrm{6}−{c} \\ $$$$\:\:\:\:\:\:\:\frac{\beta}{\gamma}\:=\:\frac{{a}}{\mathrm{6}}=\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow\:\:\:{a}=\mathrm{8} \\ $$$$\alpha+\beta=\mathrm{6}\:\:;\:\alpha\beta=\mathrm{8}\:,\:\Rightarrow\:\:\:\alpha=\mathrm{2},\mathrm{4} \\ $$$${A}\:{check}: \\ $$$${if}\:\alpha=\mathrm{2},\:{then}\:\:\:\beta=\mathrm{4}\:,\:{and}\:\gamma=\frac{\mathrm{3}}{\mathrm{4}}\beta=\mathrm{3} \\ $$$${if}\:\alpha=\mathrm{4},\:{then}\:\beta=\mathrm{2},\:{and}\:\gamma=\frac{\mathrm{3}}{\mathrm{4}}\beta=\frac{\mathrm{3}}{\mathrm{2}}\:; \\ $$$${not}\:{an}\:{integer}. \\ $$$${Hence}\:\alpha=\mathrm{2}\:.\:\:\:{c}=\alpha+\gamma\:=\mathrm{2}+\mathrm{3}=\mathrm{5}\:. \\ $$
Commented by Tinkutara last updated on 22/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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