MHB What Are the Real Solutions to This Complex Quadratic Equation?

[anemone]

Solve for the real solutions for $(3x^2-4x-1)(3x^2-4x-2)-3(3x^2-4x+5)-1=18x^2+36x-13$.

 

[mathmari]

Solve for the real solutions for $(3x^2-4x-1)(3x^2-4x-2)-3(3x^2-4x+5)-1=18x^2+36x-13$.

$(3x^2-4x-1)(3x^2-4x-2)-3(3x^2-4x+5)-1=18x^2+36x-13 \Rightarrow \\ 9x^4-12x^3-6x^2-12x^3+16x^2+8x-3x^2+4x+2-9x^2+12x-15-1=18x^2+36x-13 \Rightarrow \\ 9x^4-24x^3-2x^2+24x-14=18x^2+36x-13 \Rightarrow \\ 9x^4-24x^3-20x^2-12x-1=0$

$(3x^2+ax+b)(3x^2+cx+d)=9x^4-24x^3-20x^2-12x-1 \Rightarrow \\ 9x^4+3cx^3+3dx^2+3ax^3+acx^2+adx+3bx^2+bcx+bd=9x^4-24x^3-20x^2-12x-1 \Rightarrow \\ 9x^4+3(a+c)x^3+(3d+ac+3b)x^2+(ad+bc)x+bd=9x^4-24x^3-20x^2-12x-1 $

$3(a+c)=-24, \ \ \ 3d+ac+3b=-20, \ \ \ ad+bc=-12, \ \ \ bd=-1$

From $bd=-1$ we get: $b= \pm 1$ and $d=\mp 1$

• If $b=1,d=-1:$
  
  $3(a+c)=-24, \ \ \ -3+ac+3=-20, \ \ \ -a+c=-12$
  $-a+c=-12\Rightarrow c=a-12$
  $3(a+c)=-24 \Rightarrow 3a+3a-36=-24 \Rightarrow 6a=12 \Rightarrow a=2$
  $c=a-12 \Rightarrow c=-10$
  
  Therefore, in this case we have:
  $$(3x^2+2x+1)(3x^2-10x-1)=9x^4-24x^3-20x^2-12x-1$$
  
  $$$$
  
• If $b=-1, d=1:$
  
  $3(a+c)=-24, \ \ \ 3+ac-3=-20, \ \ \ a-c=-12$
  $a-c=-12 \Rightarrow a=c-12$
  $3(a+c)=-24 \Rightarrow 3c-36+3c=-24 \Rightarrow 6c=12 \Rightarrow c=2$
  $a=c-12 \Rightarrow a=2-12 \Rightarrow a=-10$
  
  Therefore, in this case we have:
  $$(3x^2-10x-1)(3x^2+2x+1)=9x^4-24x^3-20x^2-12x-1$$

So we have that $$9x^4-24x^3-20x^2-12x-1=(3x^2+2x+1)(3x^2-10x-1)$$

$$9x^4-24x^3-20x^2-12x-1=0 \Rightarrow (3x^2+2x+1)(3x^2-10x-1)=0 \Rightarrow \\ 3x^2+2x+1=0 \text{ or } 3x^2-10x-1=0 $$

• $3x^2+2x+1=0 \\ \Delta=4-4 \cdot 3=-8<0$
  The solutions are complex.
  $$$$
• $3x^2-10x-1=0 \\ \Delta=100-4 \cdot 3 \cdot (-1)=100+12=112>0$
  $x_{1,2}=\frac{-(-10) \pm \sqrt{\Delta}}{2 \cdot 3}=\frac{10 \pm 4 \sqrt{7}}{2 \cdot 3}=\frac{5}{3} \pm \frac{2 \sqrt{7}}{3} \in \mathbb{R}$

So the real solutions for $(3x^2-4x-1)(3x^2-4x-2)-3(3x^2-4x+5)-1=18x^2+36x-13$ are:
$$\frac{5}{3} \pm \frac{2 \sqrt{7}}{3}$$

 

[anemone]

Well done, mathmari! The answer is of course correct and thanks for participating!:)

I acknowledge that there are many different ways to approach this problem and hence, I still welcome those who wanted to take a stab at the problem differently to post your solution here. :)

 

[kaliprasad]

$(3x^2-4x-1)(3x^2-4x-2)-3(3x^2-4x+5)-1=18x^2+36x-13 \Rightarrow \\ 9x^4-12x^3-6x^2-12x^3+16x^2+8x-3x^2+4x+2-9x^2+12x-15-1=18x^2+36x-13 \Rightarrow \\ 9x^4-24x^3-2x^2+24x-14=18x^2+36x-13 \Rightarrow \\ 9x^4-24x^3-20x^2-12x-1=0$

$(3x^2+ax+b)(3x^2+cx+d)=9x^4-24x^3-20x^2-12x-1 \Rightarrow \\ 9x^4+3cx^3+3dx^2+3ax^3+acx^2+adx+3bx^2+bcx+bd=9x^4-24x^3-20x^2-12x-1 \Rightarrow \\ 9x^4+3(a+c)x^3+(3d+ac+3b)x^2+(ad+bc)x+bd=9x^4-24x^3-20x^2-12x-1 $

$3(a+c)=-24, \ \ \ 3d+ac+3b=-20, \ \ \ ad+bc=-12, \ \ \ bd=-1$

From $bd=-1$ we get: $b= \pm 1$ and $d=\mp 1$

[LIST

[*]If $b=1,d=-1:$

$3(a+c)=-24, \ \ \ -3+ac+3=-20, \ \ \ -a+c=-12$
$-a+c=-12\Rightarrow c=a-12$
$3(a+c)=-24 \Rightarrow 3a+3a-36=-24 \Rightarrow 6a=12 \Rightarrow a=2$
$c=a-12 \Rightarrow c=-10$

Therefore, in this case we have:
$$(3x^2+2x+1)(3x^2-10x-1)=9x^4-24x^3-20x^2-12x-1$$

$$$$

[*]If $b=-1, d=1:$

$3(a+c)=-24, \ \ \ 3+ac-3=-20, \ \ \ a-c=-12$
$a-c=-12 \Rightarrow a=c-12$
$3(a+c)=-24 \Rightarrow 3c-36+3c=-24 \Rightarrow 6c=12 \Rightarrow c=2$
$a=c-12 \Rightarrow a=2-12 \Rightarrow a=-10$

Therefore, in this case we have:
$$(3x^2-10x-1)(3x^2+2x+1)=9x^4-24x^3-20x^2-12x-1$$
[/LIST]

So we have that $$9x^4-24x^3-20x^2-12x-1=(3x^2+2x+1)(3x^2-10x-1)$$

$$9x^4-24x^3-20x^2-12x-1=0 \Rightarrow (3x^2+2x+1)(3x^2-10x-1)=0 \Rightarrow \\ 3x^2+2x+1=0 \text{ or } 3x^2-10x-1=0 $$

• $3x^2+2x+1=0 \\ \Delta=4-4 \cdot 3=-8<0$
  The solutions are complex.
  $$$$
• $3x^2-10x-1=0 \\ \Delta=100-4 \cdot 3 \cdot (-1)=100+12=112>0$
  $x_{1,2}=\frac{-(-10) \pm \sqrt{\Delta}}{2 \cdot 3}=\frac{10 \pm 4 \sqrt{7}}{2 \cdot 3}=\frac{5}{3} \pm \frac{2 \sqrt{7}}{3} \in \mathbb{R}$

So the real solutions for $(3x^2-4x-1)(3x^2-4x-2)-3(3x^2-4x+5)-1=18x^2+36x-13$ are:
$$\frac{5}{3} \pm \frac{2 \sqrt{7}}{3}$$

I have 2 observations

Spoiler

1) you tried $(3x^2 + ax +b)(3x^2+cx + d)$ it was a right guess but not a full proof system . It might have been $(9x^2+ax+b)(x^2+cx + d)$
2) taking b= 1 and d = -1 is same as d =1 and b = -1 they shall give same solution becuase of symetry of solution

 

[anemone]

My solution::

Spoiler

$(3x^2-4x-1)(3x^2-4x-2)-3(3x^2-4x+5)-1=18x^2+36x-13$

$(3x^2-4x)^2-6(3x^2-4x)+2-15-1=18x^2+36x-13$

At this point, I am sorely tempted to move all the terms (except the first square term) from the LHS to the RHS and simplify from there, and I get:

$(3x^2-4x)^2=36x^2+12x+1$

$(3x^2-4x)^2=(6x+1)^2$

Taking the square root on both sides we get:

$3x^2-4x=6x+1$ or $3x^2-10x-1=0$ which gives $x=\dfrac{5\pm2\sqrt{7}}{3}$.

$3x^2-4x=-(6x+1)$ or $3x^2+2x+1=0$ which gives complex roots.

Hence the solutions for the problem are $x=\dfrac{5\pm2\sqrt{7}}{3}$ and we're done.