1. **State the problem:** Solve the equation $$2x^4 - x^3 - 6x^2 - x + 2 = 0$$ using the quadratic formula. 2. **Identify the approach:** This is a quartic equation, not a quadratic. However, we can try to factor it into quadratic factors of the form $$(ax^2 + bx + c)(dx^2 + ex + f) = 0$$. 3. **Assume factorization:** Let $$ (2x^2 + px + q)(x^2 + rx + s) = 0 $$ Expanding, $$ 2x^4 + 2rx^3 + 2sx^2 + px^3 + prx^2 + psx + qx^2 + qrx + qs = 0 $$ Group like terms: $$ 2x^4 + (2r + p)x^3 + (2s + pr + q)x^2 + (ps + qr)x + qs = 0 $$ 4. **Match coefficients with original equation:** $$ 2x^4 - x^3 - 6x^2 - x + 2 = 0 $$ So, $$ 2r + p = -1 $$ $$ 2s + pr + q = -6 $$ $$ ps + qr = -1 $$ $$ qs = 2 $$ 5. **Solve for $p, q, r, s$:** Try $q=1, s=2$ (since $qs=2$), then $$ qs=1*2=2 $$ Check other equations: $$ 2r + p = -1 $$ $$ 2*2 + p*r + 1 = -6 $$ $$ p*2 + 1*r = -1 $$ Try $p=-3$, $r=2$: $$ 2*2 + (-3)*2 + 1 = 4 - 6 + 1 = -1 eq -6 $$ Try $p=1$, $r=-2$: $$ 2*(-2) + 1 = -4 + 1 = -3 eq -1 $$ Try $p=-1$, $r=0$: $$ 2*0 + (-1) = -1 $$ $$ 2s + pr + q = 2s + (-1)*0 + q = 2s + q = -6 $$ $$ ps + qr = (-1)*s + q*0 = -s = -1 ightarrow s=1 $$ $$ qs = q*1 = 2 ightarrow q=2 $$ Check $2s + q = 2*1 + 2 = 4 eq -6$ no. Try $q=2$, $s=1$: $$ qs=2*1=2 $$ $$ ps + qr = p*1 + 2*r = p + 2r = -1 $$ $$ 2r + p = -1 $$ These two are the same, so consistent. $$ 2s + pr + q = 2*1 + p*r + 2 = 2 + p*r + 2 = p*r + 4 = -6 ightarrow p*r = -10 $$ From $2r + p = -1$, express $p = -1 - 2r$. Then, $$ p*r = (-1 - 2r)r = -r - 2r^2 = -10 $$ Rearranged: $$ 2r^2 + r - 10 = 0 $$ Solve quadratic for $r$: $$ r = \frac{-1 \pm \sqrt{1 + 80}}{4} = \frac{-1 \pm 9}{4} $$ So, $$ r = 2 \text{ or } r = -\frac{5}{2} $$ If $r=2$, then $p = -1 - 2*2 = -5$. Check $ps + qr = p + 2r = -5 + 4 = -1$ correct. 6. **Factorization:** $$ (2x^2 - 5x + 2)(x^2 + 2x + 1) = 0 $$ 7. **Solve each quadratic:** - For $2x^2 - 5x + 2 = 0$, use quadratic formula: $$ x = \frac{5 \pm \sqrt{(-5)^2 - 4*2*2}}{2*2} = \frac{5 \pm \sqrt{25 - 16}}{4} = \frac{5 \pm 3}{4} $$ So, $$ x = 2 \text{ or } x = \frac{1}{2} $$ - For $x^2 + 2x + 1 = 0$, factor or use quadratic formula: $$ (x + 1)^2 = 0 \Rightarrow x = -1 $$ 8. **Final solutions:** $$ x = 2, \frac{1}{2}, -1 $$