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4-2.Quadratic Equations and Inequations
hard
समीकरण
$\log _{(x+1)}\left(2 x^{2}+7 x+5\right)+\log _{(2 x+5)}(x+1)^{2}-4=0 \text {, }$
$x > 0$ के हलों की संख्या है ..............
A
$2$
B
$4$
C
$6$
D
$1$
(JEE MAIN-2021)
Solution
$\log _{(x+1)}\left(2 x^{2}+7 x+5\right)+\log _{(2 x+5)}(x+1)^{2}-4=0$
$\log _{(x+1)}(2 x+5)(x+1)+2 \log _{(2 x+5)}(x+1)=4$
$\log _{(x+1)}(2 x+5)+1+2 \log _{(2 x+5)}(x+1)=4$
$\text { Put } \log _{(x+1)}(2 x+5)=t$
$t+\frac{2}{t}=3 \Rightarrow t^{2}-3 t+2=0$
$t=1,2$
$\log _{(x+1)}(2 x+5)=1 \quad \& \log _{(x+1)}(2 x+5)=2$
$x+1=2 x+3 \quad \& \quad 2 x+5=(x+1)^{2}$
$x=-4$ (rejected) $\quad x^{2}=4 \Rightarrow x=2,-2$ (rejected)
So, $x=2$
No. of solution $=1$
Standard 11
Mathematics
