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Sonja [21]
2 years ago
6

Solve the quadratic equation 6x2 + 1 = 5x. Show your work.

Mathematics
1 answer:
Artyom0805 [142]2 years ago
7 0

Answer:

<h2>Factors = (3x - 1) (2x - 1)</h2><h2>values :- x = 1/3 , 1/2</h2>

Step-by-step explanation:

= > 6x^2 + 1 = 5x

• Bring it in the standard form,

= > 6x^2 - 5x + 1 = 0

= > 6x^2 - (3 + 2)x + 1 = 0

= > 6x^2 - 3x - 2x + 1 = 0

• Take out common

= > 3x (2x - 1) - 1 (2x - 1) = 0

= > (3x - 1) (2x - 1) = 0...factors

= > x = 1/3 and 1/2... values of x

<h2>Hope it helps you!! </h2>

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What is a quotient of 2.5.
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Let f(x)=x²+kx+4 and g(x)=x³+x²+kx+2k, where k is a real constant.
antoniya [11.8K]

The value of k such that the graph of f and the graph of g only intersect one is equal to 2.

The value of k such that the graph of f and the graph of g only intersect one is equal to 2. According to the image attached below, functions f(x) and g(x) intersect at point (x, y) = (0, 4) for k = 2.

<h3>How to find the value of the constant k of a system of two polynomic equations</h3>

Herein we have a system formed by two <em>nonlinear</em> equations, a <em>quadratic</em> equation and a <em>cubic</em> equation. Given the constraint that both function must only intersect once, we have the following expression:

f(x) - x² - k · x = 4       (1)

g(x) - x² - k · x = x³ + 2 · k        (2)

x³ + 2 · k = 4

x³ + 2 · (k - 2) = 0

If f and g must intersect once, then the roots must of the form:

(x - r)³ = x³ + 2 · (k - 2)

x³ - 3 · r · x² + 3 · r² · x - r³ = x³ + 2 · (k - 2)

Then, the following conditions must be met: - 3 · r · x² = 0, 3 · r² · x = 0. If x may be any real number, then r must be zero and the value of k must be:

2 · (k - 2) = 0

k - 2 = 0

k = 2

Therefore, the value of k such that the graph of f and the graph of g only intersect one is equal to 2. According to the image attached below, functions f(x) and g(x) intersect at point (x, y) = (0, 4) for k = 2.

To learn more on polynomic functions: brainly.com/question/24252137

#SPJ1

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2 years ago
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