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Find the largest value of $x$ such that $3x^2 + 17x + 15 = 2x^2 + 21x + 12 - 5x^2 + 17x + 34.$

 Apr 2, 2024

Best Answer 

 #2
avatar+129732 
+1

Simplfy as

 

6x^2 - 21x -31  = 0

 

Largest  x =   [ 21 + sqrt [ 21^2 - 4*6 * -31 ]  ] / [2 * 6] ≈  4.62

 

cool cool cool

 Apr 2, 2024
 #1
avatar+1714 
0

Absolutely, I’ve been improving my problem-solving abilities in quadratic equations. Let's find the largest value of x such that: 3x2+17x+15=2x2+21x+12−5x2+17x+34

We can solve the equation by combining like terms, moving terms to one side of the equation, and using the quadratic formula.

Steps to solve: 1. Combine like terms: 3x2+17x+15=−3x2+38x+46

2. Move terms to one side: 6x2−21x−31=0

3. Use the quadratic formula: x=2a−b±b2−4ac​​ where a, b, and c are the coefficients of the quadratic equation. In this case, a = 6, b = -21, and c = -31. Substituting these values into the formula, we get:

x=2⋅6+21±(−21)2−4⋅6⋅−31​​

4. Simplify: x=1221±1185​​

Since the discriminant (the part under the radical) is positive, there are two real solutions. However, the question asks for the largest value of x. The larger solution will be the one with the positive square root.

Answer: x=121185​+21​

 Apr 2, 2024
 #2
avatar+129732 
+1
Best Answer

Simplfy as

 

6x^2 - 21x -31  = 0

 

Largest  x =   [ 21 + sqrt [ 21^2 - 4*6 * -31 ]  ] / [2 * 6] ≈  4.62

 

cool cool cool

CPhill Apr 2, 2024

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