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In Mathematics / College | 2025-07-07

What are the solutions of the equation $(2 x+3)^2+8(2 x+3)+11=0$ ? Use $u$ substitution and the quadratic formula to solve.

A. $x=\frac{-4 \pm \sqrt{5}}{2}$
B. $x=\frac{-7 \pm \sqrt{5}}{2}$
C. $x=-7$ and $x=-2$
D. $x=-1$ and $x=4$

Asked by penzoie

Answer (1)

• Perform u-substitution: Let u = 2 x + 3 , transforming the equation to u 2 + 8 u + 11 = 0 .
• Apply the quadratic formula: Use u = 2 a − b ± b 2 − 4 a c ​ ​ with a = 1 , b = 8 , and c = 11 to find u = − 4 ± 5 ​ .
• Substitute back: Replace u with 2 x + 3 to get 2 x + 3 = − 4 ± 5 ​ .
• Solve for x: Isolate x to find the solutions x = 2 − 7 ± 5 ​ ​ .
Explanation

Understanding the Problem We are given the equation ( 2 x + 3 ) 2 + 8 ( 2 x + 3 ) + 11 = 0 and asked to solve for x using u -substitution and the quadratic formula.

U-Substitution Let's use the substitution u = 2 x + 3 . This transforms the original equation into a simpler quadratic equation in terms of u : u 2 + 8 u + 11 = 0

Applying the Quadratic Formula Now we can use the quadratic formula to solve for u . The quadratic formula is given by: u = 2 a − b ± b 2 − 4 a c ​ ​ In our equation, a = 1 , b = 8 , and c = 11 . Plugging these values into the quadratic formula, we get: u = 2 ( 1 ) − 8 ± 8 2 − 4 ( 1 ) ( 11 ) ​ ​

Simplifying the Expression Let's simplify the expression under the square root: 8 2 − 4 ( 1 ) ( 11 ) = 64 − 44 = 20 So we have: u = 2 − 8 ± 20 ​ ​ We can simplify 20 ​ as 4 ⋅ 5 ​ = 2 5 ​ , so: u = 2 − 8 ± 2 5 ​ ​ Dividing both terms in the numerator by 2, we get: u = − 4 ± 5 ​

Solving for x Now we substitute back 2 x + 3 for u : 2 x + 3 = − 4 ± 5 ​ Subtract 3 from both sides: 2 x = − 7 ± 5 ​ Divide by 2 to solve for x : x = 2 − 7 ± 5 ​ ​

Final Answer Therefore, the solutions for x are x = 2 − 7 + 5 ​ ​ and x = 2 − 7 − 5 ​ ​ .


Examples
Imagine you are designing a bridge and need to calculate the stress on a particular support beam. The equation you solved is analogous to finding the points where the stress is zero, which is crucial for ensuring the bridge's stability. By using substitution and the quadratic formula, engineers can solve complex equations to determine critical parameters in structural design, ensuring safety and efficiency.

Answered by GinnyAnswer | 2025-07-07