Question

Q1.Hi !
if three distinct numbers a, b and c are in GP and the equation ax² + 2bx + c = 0 and dx² + 2ex + f = 0 have a common root , then which on of the following statements is correct .
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Answer 1

Step-by-step explanation:

a,b,c are in G.P

Therefore, b² = ac

➝ b = √(ac)

Now, We will put b = √(ac) in the given equation ax²+2bx+c = 0.

➝ ax²+2√(ac) x+c = 0

➝ (√a x)²+2.√a.√c x+ (√c)² = 0

➝ (√a x)²+2√(ac) x+ (√c)² = 0

➝ (√a x + √c)² = 0

➝ √a x + √c = 0

➝ x = -(√c ) / (√a)

Therefore, root of the equation ax²+2√(ac) x+c = 0 = -(√c ) / (√a)

Now, the equations ax² + 2bx + c = 0 and dx² + 2ex + f = 0 have a common root.

Put x = -(√c ) / (√a) in the equation dx² + 2ex + f = 0.

$\begin{gathered} \tt \rightarrow d { \bigg( \dfrac{ - \sqrt{c} }{\sqrt{a}} \bigg) }^{2} - 2e\dfrac{ \sqrt{c} }{\sqrt{a}} + f = 0 \\ \\ \tt \rightarrow d { \bigg( \dfrac{{c} }{a} \bigg) }^{} - 2e\dfrac{ \sqrt{c} }{\sqrt{a}} + f = 0\\ \\ \tt \rightarrow d { \bigg( \dfrac{{c} }{a} \bigg) }^{} - 2e\dfrac{ \sqrt{c} \times \sqrt{c} }{\sqrt{a}\times \sqrt{c}} + f = 0\\ \\ \tt \rightarrow d { \bigg( \dfrac{{c} }{a} \bigg) }^{} - 2e\dfrac{ {c} }{\sqrt{ac} } + f = 0\\ \\ \tt \rightarrow{ \dfrac{{dc} }{a} }^{} - 2\dfrac{ {ec} }{b} + f = 0\\ \\ \tt \rightarrow{ \dfrac{{dc} }{ac} }^{} - \dfrac{ {2ec} }{bc} + \dfrac{f}{c} = 0\\ \\ \tt \rightarrow{ \dfrac{{d} }{a} }^{} - \dfrac{ {2e} }{b} + \dfrac{f}{c} = 0\\ \\ \tt \rightarrow - \dfrac{ {2e} }{b} = - { \dfrac{{d} }{a} } - \dfrac{f}{c} \\ \\ \tt \rightarrow \dfrac{ {2e} }{b} = { \dfrac{{d} }{a} } + \dfrac{f}{c} \end{gathered}$

Therefore,

Therefore,d/a , e/b and f/c are in A.P

Answer 2

Step-by-step explanation:

Step-by-step explanation:

a,b,c are in G.P

Therefore, b² = ac

➝ b = √(ac)

Now, We will put b = √(ac) in the given equation ax²+2bx+c = 0.

➝ ax²+2√(ac) x+c = 0

➝ (√a x)²+2.√a.√c x+ (√c)² = 0

➝ (√a x)²+2√(ac) x+ (√c)² = 0

➝ (√a x + √c)² = 0

➝ √a x + √c = 0

➝ x = -(√c ) / (√a)

Therefore, root of the equation ax²+2√(ac) x+c = 0 = -(√c ) / (√a)

Now, the equations ax² + 2bx + c = 0 and dx² + 2ex + f = 0 have a common root.

Put x = -(√c ) / (√a) in the equation dx² + 2ex + f = 0.

\begin{gathered}\begin{gathered} \tt \rightarrow d { \bigg( \dfrac{ - \sqrt{c} }{\sqrt{a}} \bigg) }^{2} - 2e\dfrac{ \sqrt{c} }{\sqrt{a}} + f = 0 \\ \\ \tt \rightarrow d { \bigg( \dfrac{{c} }{a} \bigg) }^{} - 2e\dfrac{ \sqrt{c} }{\sqrt{a}} + f = 0\\ \\ \tt \rightarrow d { \bigg( \dfrac{{c} }{a} \bigg) }^{} - 2e\dfrac{ \sqrt{c} \times \sqrt{c} }{\sqrt{a}\times \sqrt{c}} + f = 0\\ \\ \tt \rightarrow d { \bigg( \dfrac{{c} }{a} \bigg) }^{} - 2e\dfrac{ {c} }{\sqrt{ac} } + f = 0\\ \\ \tt \rightarrow{ \dfrac{{dc} }{a} }^{} - 2\dfrac{ {ec} }{b} + f = 0\\ \\ \tt \rightarrow{ \dfrac{{dc} }{ac} }^{} - \dfrac{ {2ec} }{bc} + \dfrac{f}{c} = 0\\ \\ \tt \rightarrow{ \dfrac{{d} }{a} }^{} - \dfrac{ {2e} }{b} + \dfrac{f}{c} = 0\\ \\ \tt \rightarrow - \dfrac{ {2e} }{b} = - { \dfrac{{d} }{a} } - \dfrac{f}{c} \\ \\ \tt \rightarrow \dfrac{ {2e} }{b} = { \dfrac{{d} }{a} } + \dfrac{f}{c} \end{gathered} \end{gathered}

→d(

a

−

c

)

2

−2e

a

c

+f=0

→d(

a

c

)

−2e

a

c

+f=0

→d(

a

c

)

−2e

a

×

c

c

×

c

+f=0

→d(

a

c

)

−2e

ac

c

+f=0

→

a

dc

−2

b

ec

+f=0

→

ac

dc

−

bc

2ec

+

c

f

=0

→

a

d

−

b

2e

+

c

f

=0

→−

b

2e

=−

a

d

−

c

f

→

b

2e

=

a

d

+

c

f

Therefore,

Therefore,d/a , e/b and f/c are in A.P