Vieta’s Formula

It’s easy to get $\Delta=b^2-4ac=3^2-4 \times 1 \times 1>0$ . Obviously, $x_1 \neq x_2$ . You can certainly solve the equation, but we can substitute $x$ by $x_1$ and $x_2$ . $\begin{aligned} \because x^2+3x+1 &=0. \\ \therefore {x_1}^2+3 x_1 +1 &=0. \\ \therefore {x_1}^2 + 2 x_1 &=-(x_1 + 1). \\ \because x^2+3x+1 &=0. \\ \therefore{x_2}^2+3 x_2 +1 &=0. \\ \therefore{x_2}^2 + 4 x_2 + 2 &=x_2 + 1 \end{aligned}$ $\because a=1, b=3, c=1$ $\begin{aligned} \therefore \left( {x_1}^2 +2 x_1 \right)\left( {x_2}^2 +4 x_2 + 2\right) &=-\left[(x_1 + 1)(x_2 + 1)\right]. \\ &=-\left[ x_1 x_2 + ( x_1 + x_2 ) + 1 \right]. \\ &=-\left( \frac{c}{a}-\frac{b}{a} +1 \right). \\ &=-\left( \frac{1}{1}-\frac{3}{1}+ 1 \right). \\ &=-(1-3+1). \\ &=1 \end{aligned}$ $\therefore\left( {x_1}^2 +2 x_1 \right)\left( {x_2}^2 +4 x_2 + 2\right)=1.$

$\begin{aligned} X & = (x_1^2 + 2x_1)(x_2^2 + 4x_2 + 2) \\ & = (\blue{x_1^2 + 3x_1+1} - x_1 -1)(\blue{x_2^2+3x_2+1} + x_2 + 1) & \blue{\text{Note that }x^2+3x+1 = 0} \\ & = (\blue{0} - x_1 -1)(\blue{0} + x_2 + 1) \\ & = -(x_1+1)(x_2+1) \\ & = - (x_1x_2 + x_1 + x_2 +1) & \small \blue{\text{By Vieta's formula }x_1x_2 = 1 \text{ and }x_1+x_2 = - 3} \\ & = - (1-3+1) = \boxed 1 \end{aligned}$

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Reference: Vieta's formula

We have $x_1+x_2=-3,x_1x_2=1$

Expanding the given expression and using the relations we get the given expression as

$2x_1^2+8x_1+9+2x_2=2(x_1^2+3x_1+1)+2x_1+2x_2+7$

$=2(x_1+x_2)+7=2(-3)+7=\boxed 1$ .