Question

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

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.

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

Therefore,

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

Answer 2

Answer:

option (b) d/a , e/b and f/x are in AP

Step-by-step explanation:

Given:- , Three distinct numbers a, b and c are in GP. b² = ac ______(i)

and the given quadratic equations

ax² + 2bx + x = 0 ______(ii)

dx² + 2ex + f = 0 _____(iii)

for quadratic equation (ii)

the discriminant D = (2b)² -4ac

= 4(b²-ac) = 0 [from eq (i) ]

=> Quadratic eq . (ii) have equal roots, and it is equal to x = -b/a ,and it is given that quadratic

eqs , (ii) and (iii) have a common root, so

d ( -b/a)² + 2e ( - b/a ) + f = 0

=> db² - 2eab + af² = 0 [∵b² = ac]

=> d(ac ) -2eab + a²f = 0 [∵ a≠0]

=> dc - 2eb + af

=> 2eb = dc + af

=> 2e/b = dc/b² + af/b² (dividing each term by b²)

=> 2e/b = d/a + f/c

[∵ 2b = a+ c is known as in AP ]

similarly, d/a , e/b, f/c Are in AP

∴ oprtion (b) is correct