The sum of squares of all the real solutions of the equation _(x+1)(2x^2+5x+3)=4- _(2x+3)(…

Question

The sum of squares of all the real solutions of the equation

${\log }_{(x+1)}(2{x}^{2}+5x+3)=4-{\log }_{(2x+3)}({x}^{2}+2x+1)$ is equal to $____$ .

Answer: 2

Accepted Answer

${\log }_{(x+1)}(2{x}^{2}+5x+3)=4-{\log }_{(2x+3)}({x}^{2}+2x+1)$ $\Rightarrow$ ${\log }_{(x+1)}(x+1)(2x+3)=4-{\log }_{(2x+3)}(x+1{)}^{2}$ $\Rightarrow$ ${\log }_{(x+1)}(x+1)+{\log }_{(x+1)}(2x+3)=4-2{\log }_{(2x+3)}(x+1)$ $\Rightarrow$ $1+\frac{1}{{\log }_{(2x+3)}(x+1)}=4-2{\log }_{(2x+3)}(x+1)$ Substitute ${\log }_{(2x+3)}(x+1)=t$ $ herefore $ $1+\frac{1}{t}=4-2t\Rightarrow 2{t}^{2}-3t+1=0$ $\Rightarrow (t-1)(2t-1)=0\Rightarrow t={1,\frac{1}{2}}$ For t = 1, ${\log }_{(2x+3)}(x+1)=1\Rightarrow 2x+3=x+1\Rightarrow x=-2$ For t = $\frac{1}{2}$, ${\log }_{(2x+3)}(x+1)=\frac{1}{2}\Rightarrow (x+1{)}^{2}=2x+3\Rightarrow {x}^{2}=2\Rightarrow x=\pm \sqrt{2}$ j xj x + 1 Base 2x + 3 Base 2x2 + 5x + 3 Argument x2 + 2x + 1 Argument Remarks Verdict -2 -1 < 0 -1 < 0 > 0 > 0 1st and 2nd base -ve Invalid -\sqrt 2 1 - \sqrt 2 < 0 3 - 2\sqrt 2 > 0 7 - 5\sqrt 2 < 0 3 - 2\sqrt 2 > 0 1st base and 1st argument -ve Invalid \sqrt 2 1 + \sqrt 2 > 0 3 + 2\sqrt 2 > 0 7 + 5\sqrt 2 > 0 3 + 2\sqrt 2 > 0 All bases and arguments +ve Valid $\sum _{j=1}^{3}{[{x}_{j}]}^{2}=0+0+(\sqrt{2}{)}^{2}=2$

Solutions · 1