If three distinct numbers a, b, c are in G.P. and the
equations ax2
+ 2bx + c = 0 and
dx2
+ 2ex + f = 0 have a common root, then
which one of the following statements is
correct?
Solutions · 1
Official solution
Given, a, b, c are in G.P.
\[ \therefore \] b2 = ac
In this equation ax2
+ 2bx + c = 0,
Discrimant, D = 4b2 - 4ac
= 4ac - 4ac
= 0
Discrimant = 0 meand roots of the equation are equal.
Let both the roots of the equation = \[\alpha \]
\[ \therefore \] 2\[\alpha \] = \[ - {{2b} \over a}\]
\[ \Rightarrow \] \[\alpha \] = \[ - {b \over a}\]
As both the
equations ax2
+ 2bx + c = 0 and
dx2
+ 2ex + f = 0 have a common root,
so \[ - {b \over a}\] is also root of the equation dx2
+ 2ex + f = 0.
\[ \therefore \] \[ - {b \over a}\] satisfy the equation dx2
+ 2ex + f = 0.
\[ \therefore \] \[d{\left( { - {b \over a}} \right)^2} + 2e\left( { - {b \over a}} \right) + f = 0\]
\[ \Rightarrow \] db2−2aeb+fa2=0
\[ \Rightarrow \] dac−2aeb+fa2=0
\[ \Rightarrow \] dc−2eb+fa=0
\[ \Rightarrow \] \[{{dc} \over {ac}} - {{2eb} \over {ac}} + {{fa} \over {ac}} = 0\]
\[ \Rightarrow \] \[{{dc} \over {ac}} - {{2eb} \over {{b^2}}} + {{fa} \over {ac}} = 0\]
\[ \Rightarrow \] \[{d \over a} - {{2e} \over b} + {f \over c} = 0\]
\[ \Rightarrow \] \[{{2e} \over b} = {d \over a} + {f \over c}\]
\[ \therefore \]
\[d \over a\], \[e \over b\], \[f \over c\] are in A.P.