If a,b,c are in G.P then the equations ax^2 + 2bx + c = 0,dx^2 + 2ex + f = 0 have a common root if (d / a),(e / b),(f / c) are in

(a) : a, b, c are in G.P, b=sqrt(a c) \ Therefore a x^2+2 sqrt(a c) x+c=0 (sqrt(a) x+sqrt(c))^2 =0 Therefore x=-sqrt((c / a)) x=-sqrt((c / a)) must satisfy the second equation \ Therefore d * (c / a)-2 e sqrt((c / a))+f=0 dividing by c, (d / a)-frac2 esqrt(a c)+(f / c)=0 \ Therefore (2 e / b)=(d / a)+(f / c) (Because b =sqrt(a c)) Therefore (d / a), (e / b), (f / c) are in A.P.

If a,b,c are in G.P then the equations ax^{2} + 2bx +... | Sequence & Series (Mathematics)

(a) : $a, b, c$ are in G.P, $b=\sqrt{a c} \\ \therefore a x^2+2 \sqrt{a c} x+c=0 (\sqrt{a} x+\sqrt{c})^2 \\=0 \quad \therefore x=-\sqrt{\frac{c}{a}} x=-\sqrt{\frac{c}{a}}\\$ must satisfy the second equation $\\ \therefore d \cdot \frac{c}{a}-2 e \sqrt{\frac{c}{a}}+f=0\\$ dividing by $c, \frac{d}{a}-\frac{2 e}{\sqrt{a c}}+\frac{f}{c}=0 \\ \therefore \frac{2 e}{b}=\frac{d}{a}+\frac{f}{c} \\(\because b =\sqrt{a c}) \therefore \frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.

Explanation

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