Work out the solutions to 4x = 3x^2 - 7x + 9, giving your answers to 3 significant figures.

Step-by-step solution

1. 1

   Collect everything on the side where the squared term is positive. Subtract $4x$ from both sides so the equation reads $=0$ with a positive leading coefficient:

   $0=3x^{2}-7x-4x+9\quad\Longrightarrow\quad 3x^{2}-11x+9=0.$

2. 2

   Check whether it factors before reaching for the formula. You would need two integers multiplying to $3\cdot 9=27$ and adding to $-11$; the pairs are $(-1,-27)$ and $(-3,-9)$, summing to $-28$ and $-12$. Neither works, so the roots are irrational and the quadratic formula is the right tool.

3. 3

   Identify the coefficients and compute the discriminant. With $a=3$, $b=-11$, $c=9$,

   $b^{2}-4ac=(-11)^{2}-4(3)(9)=121-108=13.$

   Since $13>0$ there are two distinct real roots, and since $13$ is not a perfect square they are irrational — consistent with the instruction to round.

4. 4

   Apply the quadratic formula.

   $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{11\pm\sqrt{13}}{6}.$

5. 5

   Evaluate and round to 3 significant figures. With $\sqrt{13}=3.605551\ldots$,

   $x=\frac{11+3.605551}{6}=2.434258\ldots\approx 2.43,\qquad x=\frac{11-3.605551}{6}=1.232408\ldots\approx 1.23.$

   Round only at this final step — rounding $\sqrt{13}$ early can shift the third significant figure.

6. 6

   Verify with the exact roots. Vieta's relations give $x_1+x_2=\tfrac{11}{3}\approx 3.6667$ and $x_1x_2=\tfrac{9}{3}=3$; the computed values sum to $3.66666$... which is $\tfrac{22}{6}$, and indeed $2.434258+1.232408=3.666666=\tfrac{11}{3}$, matching $-b/a$. Their product is $2.434258\times 1.232408=3.0000$, matching $c/a=3$.