Solve 0.02x^2 + 0.06x + 0.18 = 2.91. Give the exact solutions in surd form and their decimal approximations.

Step-by-step solution

1. 1

   Put the equation in standard form first. The quadratic formula only applies to $ax^{2}+bx+c=0$, so subtract $2.91$ from both sides before anything else:

   $0.02x^{2}+0.06x+0.18-2.91=0\;\Longrightarrow\;0.02x^{2}+0.06x-2.73=0.$

   Note the constant becomes negative, which already tells you the two roots will have opposite signs (their product $c/a$ is negative).

2. 2

   Clear the decimals by multiplying through by 100. Decimal coefficients make the discriminant painful to compute by hand and invite rounding slips. Multiplying every term by $100$ is an equivalence, not an approximation:

   $2x^{2}+6x-273=0.$

   Check each term: $0.02\times100=2$, $0.06\times100=6$, $-2.73\times100=-273$. All three are now integers.

3. 3

   Apply the quadratic formula with $a=2$, $b=6$, $c=-273$. Substituting,

   $x=\frac{-6\pm\sqrt{6^{2}-4(2)(-273)}}{2(2)}=\frac{-6\pm\sqrt{36+2184}}{4}=\frac{-6\pm\sqrt{2220}}{4}.$

   The sign trap is $-4ac=-4(2)(-273)=+2184$: two minus signs inside make the discriminant larger, not smaller. Since $2220>0$ there are two distinct real roots.

4. 4

   Simplify the surd and cancel the common factor. Pull the largest square factor out of $2220$:

   $2220=4\cdot 555,\qquad \sqrt{2220}=2\sqrt{555},$

   and $555=3\cdot 5\cdot 37$ has no square factor, so this is fully simplified. Then

   $x=\frac{-6\pm 2\sqrt{555}}{4}=\frac{-3\pm\sqrt{555}}{2}.$

   Cancelling the $2$ from all three parts of the fraction is what turns an ugly answer into a clean one.

5. 5

   Approximate and check both roots in the original equation. With $\sqrt{555}\approx 23.5584$,

   $x\approx\frac{-3+23.5584}{2}\approx 10.2792,\qquad x\approx\frac{-3-23.5584}{2}\approx-13.2792.$

   Substituting $x=10.2792$ back: $0.02(105.66)+0.06(10.2792)+0.18=2.1132+0.6168+0.18=2.9100$ ✓. The same check on $x=-13.2792$ also returns $2.91$ ✓. The two roots are symmetric about the vertex $x=-1.5$, as they must be.