#### Question

Solve (4x+3)^2(2x+1)(x+1)=9

Question

Solve (4x+3)^2(2x+1)(x+1)=9

This is a quartic equation. It would be complex to expand the equation to its general form and solve in conventional methods.

Here let’s use substitution method to find the solutions.

Let

t=4x+3

(1)

Then,

(4x+3)^2(2x+1)(x+1)

=t^2(t-1)(t+1)

=72

Expanding the equation in terms of t gives biquadratic equation,

t^4-t^2-72=0

Factorizing the LHS

Then we get 4 solutions for t

(t^2-9)(t^2+8)=0

\begin{cases} t_1=3 \\ t_2=-3 \\ t_3=2\sqrt{2}i \\ t_4=-2\sqrt{2}i \end{cases}

Substitute the values of t to (1), then we obtain 4 solutions of the original equation.

\begin{cases} x_1=0 \\ x_2=-\dfrac{3}{2} \\ x_3=-\dfrac{3}{4}+\dfrac{\sqrt{2}i }{2} \\ x_4=-\dfrac{3}{4}-\dfrac{\sqrt{2}i }{2} \end{cases}