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

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

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

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

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