Let a,b,c,d ∈R. If the equation 2bx2+3cx−d=0 and 2ax2+3bx+4c=0 have a common root and 4bc+adk(b2−ac)=bd+4c24bc+ad, then k=

Let α be the common root of2bx2+3cx−d=0 and 2ax2+3bx+4c=0⇒ 2bα2+3cα−d=0 and 2aα2+3bα+4c=0by solving α2 α 13c−d2b3c2b4c2a3b⇒ α212c2+3bc=α−2ad−8bc=16b2−6ac⇒ α23(bc+4c2)=α−2(4bc+ad)=16(b2−ac)α2=3(bc+4c2)b(b2−ac),α=−2(4bc+ad)6(b2−ac)α2=bc+4c22(b2−ac),α=−(4bc+ad)3(b2−ac)(4bc+ad)29(b2−ac)2= abc+4c22(b2−ac)⇒ (4bc+ad)29(b2−ac)= bc+4c22⇒ 4bc+ad(92)(b2−ac)=bc+4c24bc+ad∴ k= 92

Let a,b,c,d ∈R. If the equation 2bx2+3cx−d=0 and 2ax2+3bx+4c=0 have a common root and 4bc+adk(b2−ac)=bd+4c24bc+ad, then k=

Let α be the common root of2bx2+3cx−d=0 and 2ax2+3bx+4c=0⇒ 2bα2+3cα−d=0 and 2aα2+3bα+4c=0by solving α2 α 13c−d2b3c2b4c2a3b⇒ α212c2+3bc=α−2ad−8bc=16b2−6ac⇒ α23(bc+4c2)=α−2(4bc+ad)=16(b2−ac)α2=3(bc+4c2)b(b2−ac),α=−2(4bc+ad)6(b2−ac)α2=bc+4c22(b2−ac),α=−(4bc+ad)3(b2−ac)(4bc+ad)29(b2−ac)2= abc+4c22(b2−ac)⇒ (4bc+ad)29(b2−ac)= bc+4c22⇒ 4bc+ad(92)(b2−ac)=bc+4c24bc+ad∴ k= 92

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$\begin{array}{l} L e t a, b, c, d \in R . I f t h e e q u a t i o n 2 b x^{2} + 3 c x - d = 0 a n d 2 a x^{2} + 3 b x + 4 c = 0 h a v e a c o m m o n r o o t a n d \\ \frac{4 b c + a d}{k (b^{2} - a c)} = \frac{b d + 4 c^{2}}{4 b c + a d}, t h e n k = \end{array}$

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$\frac{9}{2}$

$\frac{2}{9}$

$\frac{1}{9}$

$\frac{1}{3}$

answer is A.

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Detailed Solution

$\begin{array}{l} L e t α b e t h e c o m m o n r o o t o f \\ 2 b x^{2} + 3 c x - d = 0 a n d 2 a x^{2} + 3 b x + 4 c = 0 \\ \Rightarrow 2 b α^{2} + 3 c α - d = 0 a n d 2 a α^{2} + 3 b α + 4 c = 0 \\ b y s o l v i n g \\ α^{2} α 1 \\ \begin{matrix} 3 c & - d & 2 b & 3 c \\ 2 b & 4 c & 2 a & 3 b \end{matrix} \\ \Rightarrow \frac{α^{2}}{12 c^{2} + 3 b c} = \frac{α}{- 2 a d - 8 b c} = \frac{1}{6 b^{2} - 6 a c} \\ \Rightarrow \frac{α^{2}}{3 (b c + 4 c^{2})} = \frac{α}{- 2 (4 b c + a d)} = \frac{1}{6 (b^{2} - a c)} \\ α^{2} = \frac{3 (b c + 4 c^{2})}{b (b^{2} - a c)}, α = \frac{- 2 (4 b c + a d)}{6 (b^{2} - a c)} \\ α^{2} = \frac{b c + 4 c^{2}}{2 (b^{2} - a c)}, α = \frac{- (4 b c + a d)}{3 (b^{2} - a c)} \\ \frac{{(4 b c + a d)}^{2}}{9 {(b^{2} - a c)}^{2}} = \frac{a b c + 4 c^{2}}{2 (b^{2} - a c)} \\ \Rightarrow \frac{{(4 b c + a d)}^{2}}{9 (b^{2} - a c)} = \frac{b c + 4 c^{2}}{2} \\ \Rightarrow \frac{4 b c + a d}{(\frac{9}{2}) (b^{2} - a c)} = \frac{b c + 4 c^{2}}{4 b c + a d} \\ ∴ k = \frac{9}{2} \end{array}$