The problem asks us to find the values of $x$ that satisfy the equation $5x = 3x^2 - 9x + 12$. We need to give the solutions to 3 significant figures.

AlgebraQuadratic EquationsQuadratic FormulaSignificant FiguresAlgebraic Manipulation
2025/5/7

1. Problem Description

The problem asks us to find the values of xx that satisfy the equation 5x=3x29x+125x = 3x^2 - 9x + 12. We need to give the solutions to 3 significant figures.

2. Solution Steps

First, we rewrite the equation in the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0:
3x29x+125x=03x^2 - 9x + 12 - 5x = 0
3x214x+12=03x^2 - 14x + 12 = 0
Now, we can use the quadratic formula to solve for xx:
x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
In this case, a=3a = 3, b=14b = -14, and c=12c = 12. Substituting these values into the quadratic formula:
x=(14)±(14)24(3)(12)2(3)x = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(3)(12)}}{2(3)}
x=14±1961446x = \frac{14 \pm \sqrt{196 - 144}}{6}
x=14±526x = \frac{14 \pm \sqrt{52}}{6}
x=14±2136x = \frac{14 \pm 2\sqrt{13}}{6}
x=7±133x = \frac{7 \pm \sqrt{13}}{3}
So, the two possible values for xx are:
x1=7+133x_1 = \frac{7 + \sqrt{13}}{3} and x2=7133x_2 = \frac{7 - \sqrt{13}}{3}.
Now, we approximate these values to 3 significant figures:
x1=7+1337+3.60555310.6055533.535183.54x_1 = \frac{7 + \sqrt{13}}{3} \approx \frac{7 + 3.60555}{3} \approx \frac{10.60555}{3} \approx 3.53518 \approx 3.54
x2=713373.6055533.3944531.131481.13x_2 = \frac{7 - \sqrt{13}}{3} \approx \frac{7 - 3.60555}{3} \approx \frac{3.39445}{3} \approx 1.13148 \approx 1.13

3. Final Answer

The values of xx to 3 significant figures are x=3.54x = 3.54 and x=1.13x = 1.13.

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