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64x^2-(16a+4b)x+ab=0 In the given equation, a and b are positive constants. The sum of the solutions to the given equation is k(4a+b) , where k is a constant. What is the value of k ? a. -1/8 b. 1/8 c. 1/4 d. -1/4

User Helle
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Final answer:

The sum of the solutions to the given equation is k(4a+b), where k is a constant. The value of k is 1/8.

Step-by-step explanation:

The given equation is a quadratic equation of the form ax² + bx + c = 0, with a = 64, b = -(16a+4b), and c = ab. We need to find the sum of the solutions to the equation in terms of the constants a and b. To do this, we can use the quadratic formula.

The quadratic formula is: x = (-b ± √(b² - 4ac)) / (2a). Substituting the values of a, b, and c into the formula, we get:

x = (-(16a+4b) ± √((16a+4b)² - 4(64)(ab))) / (2*64)

Simplifying further, we have:

x = (-(16a+4b) ± √((16a+4b)² - 256ab)) / 128

The sum of the solutions is given by the expression k(4a+b), where k is a constant. Comparing this expression with the above equation, we can see that k = 1/8. Therefore, the value of k is b. 1/8.

User Lasso
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