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solve by completing the square. 4u² + 112u + 44 = 0 write your answers …

Question

solve by completing the square.
4u² + 112u + 44 = 0
write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.
u = or u =

Explanation:

Step1: Divide by 4

Divide the entire equation \(4u^{2}+112u + 44=0\) by 4. We get \(u^{2}+28u + 11 = 0\). Then, rewrite it as \(u^{2}+28u=- 11\).

Step2: Complete the square

To complete the square for the \(u\) terms, take half of the coefficient of \(u\) (which is 28), square it. Half of 28 is 14, and \(14^{2}=196\). Add 196 to both sides of the equation: \(u^{2}+28u + 196=-11 + 196\).

Step3: Factor and simplify

The left - hand side factors as \((u + 14)^{2}\), and the right - hand side is \(185\). So, \((u + 14)^{2}=185\).

Step4: Solve for u

Take the square root of both sides: \(u+14=\pm\sqrt{185}\). Then, solve for \(u\): \(u=-14\pm\sqrt{185}\). Calculate the numerical values: \(\sqrt{185}\approx13.60\). So, \(u=-14 + 13.60=-0.40\) or \(u=-14-13.60=-27.60\) (rounded to the nearest hundredth). Or in exact form, \(u=-14+\sqrt{185}\) or \(u=-14 - \sqrt{185}\). If we simplify \(\sqrt{185}\) is irrational, and as a fraction - like form (but it's better to present decimal or exact radical form). Let's calculate the exact decimal values: \(\sqrt{185}\approx13.60147\), so \(u=-14 + 13.60147\approx - 0.3985\approx - 0.40\) and \(u=-14-13.60147=-27.60147\approx - 27.60\). Also, if we want to present as fractions, from the quadratic formula (alternatively), for \(au^{2}+bu + c = 0\), \(u=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 4\), \(b = 112\), \(c = 44\). \(u=\frac{-112\pm\sqrt{112^{2}-4\times4\times44}}{2\times4}=\frac{-112\pm\sqrt{12544 - 704}}{8}=\frac{-112\pm\sqrt{11840}}{8}=\frac{-112\pm4\sqrt{740}}{8}=\frac{-28\pm\sqrt{740}}{2}=\frac{-28\pm2\sqrt{185}}{2}=-14\pm\sqrt{185}\), which matches our previous result.

Answer:

\(u\approx - 0.40\) or \(u\approx - 27.60\) (or in exact form \(u=-14+\sqrt{185}\) or \(u=-14 - \sqrt{185}\))