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liraira [26]
2 years ago
14

Consider the equation 2(ax+3)=4x−3(4x+8). For what value of a is there no solution to the equation?

Mathematics
1 answer:
erastovalidia [21]2 years ago
5 0

Answer:

the value of a such that  there is no solution to the equation is a = -4.

Step-by-step explanation:

Let first simplify the expression presented on statement. The equation has no solution if and only if a is eliminated in the process and an absurd is the result (i.e. 0 = 7).

2\cdot (a\cdot x + 3) = 4\cdot x -3\cdot (4\cdot x + 8)

2\cdot a\cdot x +6 = 4\cdot x -12\cdot x -24

2\cdot a \cdot x+6=-8\cdot x-24

2\cdot a\cdot x +8\cdot x = -30

2\cdot (a+4)\cdot x = -30

(a+4)\cdot x = -15

To obtain an absurd, we need that a+4 = 0. Hence, the value of a such that  there is no solution to the equation is:

a = -4

Let prove the certainty of the result. We find that an absurd exist: (a = -4)

0 = -15

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2 years ago
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3 0
3 years ago
I NEED HELP!
Vinvika [58]

Answer:

Step-by-step explanation:

First of all you have to put the zeros into factor form.

y1 = (x + 2)(x - 5)

Notice that the x changes sigh. You say that the zeros are -2 and  5. To get y1 to go to zero, you must make x the opposite sign of what you are given.

Now you have to expand the factored form to get the standard form

y = x^2 + 2x - 5x - 10

y = x^2 - 3x - 10

That's not the right answer.

f(0) = 30 but you have - 10 at the end so you have to multiply the factored form by - 3

-3(f(x)) = -3(x + 2)(x - 5)

-3(f(x)) = -3(x^2 - 3x - 10)

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f(0) = -3(0)^2 + 9(0) + 30

f(0) = 30

The quadratic still has roots of (x +2)(x - 5) but that 3 makes the y intercept = 30.

See the graph below.

6 0
3 years ago
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