Question

Solve the equation g(x) = 2k(x) algebraically for x, to the nearest tenth. given

g(x)=2x^+3x+10
k(x)=2x+16

Answer 1

To solve the equation g(x) = 2k(x) algebraically for x, substitute the given expressions for g(x) and k(x) into the equation and simplify it. Then, use the quadratic formula to solve for x. In this case, the equation has no real solutions.

Step-by-step explanation:

To solve the equation g(x) = 2k(x) algebraically for x, we substitute the given expressions for g(x) and k(x) into the equation. We get:

2x^2 + 3x + 10 = 2(2x + 16)

Expanding and rearranging the equation:

2x^2 + 3x + 10 = 4x + 32

Combining like terms:

2x^2 - x + 22 = 0

We can solve this equation using the quadratic formula. Plugging the values into the formula:

x = (-(-1) ± √((-1)^2 - 4(2)(22))) / (2(2))

Simplifying further:

x = (1 ± √(1 - 176)) / 4

Since the square root value is negative, the equation has no real solutions.

Answer 2

Using the degree of freedom rule, we can solve three unknown variables if and only if the number of independent equations is equal to 3. Thus the number of equations should be equal to the number of variables. We can use substitution to find x.

g(x) = 2k(x) (1)
g(x)=2x^+3x+10 (2)
k(x)=2x+16 (3)

we substitute 2 to 1 and also 3 to 1. The resulting function hence becomes:
2x^+3x+10 = 2 * (2x +16)
Simplifying the equation on the right.
2x^+3x+10 = 4x +32
we group then the like terms on one side. That is,
2x^+3x - 4x+10 -32 = 0
2x^2 - x - 22 = 0
The factors using the quadratic equation are

x1 ==1/4+1/4√177
x2 ==1/4-1/4√177