If-the-roots-of-the-equation-ax-2-bx-5c-0-are-in-the-ratio-of-4-5-then-4b-2-81ac-

Question Number 30183 by Surajitzzz last updated on 17/Feb/18

$$\mathrm{If}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:{ax}^{\mathrm{2}} −{bx}+\mathrm{5}{c}=\mathrm{0} \\ $$$$\mathrm{are}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{4}:\:\mathrm{5},\:\mathrm{then}\:\mathrm{4}{b}^{\mathrm{2}} =\mathrm{81}{ac}. \\ $$

Answered by Rasheed.Sindhi last updated on 18/Feb/18

Let the roots are 4k & 5k Sum of the roots =(b/a)=(4+5)k=9k ⇒k=(b/(9a)) Product of the roots =((5c)/a)=(4k)(5k)=20k^2 =20((b/(9a)))^2 =((20b^2 )/(81a^2 )) ((5c)/a)=((20b^2 )/(81a^2 )) c=((4b^2 )/(81a)) 4b^2 =81ac

$$\mathrm{Let}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{4k}\:\&\:\mathrm{5k} \\ $$$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\:=\frac{\mathrm{b}}{\mathrm{a}}=\left(\mathrm{4}+\mathrm{5}\right)\mathrm{k}=\mathrm{9k} \\ $$$$\:\:\:\:\:\:\Rightarrow\mathrm{k}=\frac{\mathrm{b}}{\mathrm{9a}} \\ $$$$\mathrm{Product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:=\frac{\mathrm{5c}}{\mathrm{a}}=\left(\mathrm{4k}\right)\left(\mathrm{5k}\right)=\mathrm{20k}^{\mathrm{2}} \\ $$$$\:\:\:\:=\mathrm{20}\left(\frac{\mathrm{b}}{\mathrm{9a}}\right)^{\mathrm{2}} =\frac{\mathrm{20b}^{\mathrm{2}} }{\mathrm{81a}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{5c}}{\mathrm{a}}=\frac{\mathrm{20b}^{\mathrm{2}} }{\mathrm{81a}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{c}=\frac{\mathrm{4b}^{\mathrm{2}} }{\mathrm{81a}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{4b}^{\mathrm{2}} =\mathrm{81ac} \\ $$

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