Algebra · real student question

Solve for Vout: 1.2777 + 1.1837 x 0.999Vout x 0.000288 x (284365 + (0.999Vout) squared / 2) - 1.182677 x Vout x 0.000288 x (285105 + Vout squared / 2) = 0.

Question

Solve for VoutV_{out}: 1.2777+1.1837(0.999Vout)(0.000288)(284365+(0.999Vout)22)−1.182677 Vout(0.000288)(285105+Vout22)=0.1.2777 + 1.1837(0.999V_{out})(0.000288)\left(284365+\frac{(0.999V_{out})^2}{2}\right) - 1.182677\,V_{out}(0.000288)\left(285105+\frac{V_{out}^2}{2}\right)=0.

Step-by-step solution

  1. See the structure: it is a cubic in Vout. Each bracket contributes a constant term and a V2V^2 term, and each is multiplied by VV, so the equation has the form A+BV+CV3=0A + BV + CV^3 = 0 with no V2V^2 term. There is no need for a general cubic formula - the numbers decide everything.

  2. Collect the first group's coefficients exactly. 1.1837×0.999×0.000288=0.00034056469441.1837 \times 0.999 \times 0.000288 = 0.0003405646944 (not 0.000340740.00034074 - carry all digits). Multiplying by the bracket: the linear part is 0.0003405646944×284365=96.8446793230.0003405646944 \times 284365 = 96.844679323 and the cubic part is 0.0003405646944×0.99922=0.000170002280.0003405646944 \times \tfrac{0.999^2}{2} = 0.00017000228.

  3. Collect the second group's coefficients. 1.182677×0.000288=0.0003406109761.182677\times0.000288 = 0.000340610976, giving a linear part 0.000340610976×285105=97.1098923120.000340610976\times285105 = 97.109892312 and a cubic part 0.0003406109762=0.000170305488\tfrac{0.000340610976}{2} = 0.000170305488.

  4. Subtract - and respect the cancellation. B=96.844679323−97.109892312=−0.265212989,C=0.000170002280−0.000170305488=−3.635352×10−7.B = 96.844679323-97.109892312 = -0.265212989,\qquad C = 0.000170002280-0.000170305488 = -3.635352\times10^{-7}. Both differences are catastrophic cancellations: BB loses three significant digits and CC loses four, so rounding either factor to six digits would destroy the answer. Working with exact fractions is the only safe route here.

  5. Solve the resulting cubic. 1.2777−0.265212989 V−3.635352×10−7V3=0.1.2777-0.265212989\,V-3.635352\times10^{-7}V^3 = 0. The cubic term is negligible near the root, so the linear estimate is V≈1.27770.265213=4.81753V\approx\tfrac{1.2777}{0.265213}=4.81753; one Newton correction (or bisection on [0,20][0,20]) gives Vout≈4.81748.V_{out} \approx 4.81748.

  6. Verify and note the sensitivity. At V=4.81748V=4.81748: −0.265212989(4.81748)=−1.277659-0.265212989(4.81748) = -1.277659 and −3.635352×10−7(4.81748)3=−0.0000406-3.635352\times10^{-7}(4.81748)^3 = -0.0000406, and 1.2777−1.277659−0.0000406≈01.2777-1.277659-0.0000406 \approx 0. Using the slightly-off coefficient −0.20276-0.20276 instead would return 6.30146.3014 - a 31%31\% error from a fifth-digit slip, which is exactly why the cancellation must be handled exactly.

Answer

Vout≈4.81748V_{out} \approx 4.81748

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