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3 years ago

Which of the following is the solution for 9<=1-4x<21

Mathematics

1 answer:

frozen [14]3 years ago

5 0

Answer:

-5 < x <=-2

Step-by-step explanation:

9<=1-4x<21

Subtract 1 from each side

9 -1<=1-1-4x<21-1

8 <= -4x < 20

Divide each side by -4. That means to flip all inequalities

8/-4 >= -4x/-4 >20/-4

-2 >=x > -5

-5 < x <=-2

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When the sum of \, 528 \, and three times a positive number is subtracted from the square of the number, the result is \, 120. F

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Then, we have to subtract this number from $x^2$ , so we have

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This is a quadratic equation, i.e. an equation like $ax^2+bx+c=0$ . These equation can be solved - assuming they have a solution - with the following formula

$x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

If you plug the values from your equation, you have

$x_{1,2} = \dfrac{3\pm\sqrt{9-4\cdot(-648)}}{2} = \dfrac{3\pm\sqrt{9+2592}}{2} = \dfrac{3\pm\sqrt{2601}}{2} = \dfrac{3\pm51}{2}$

So, the two solutions would be

$x = \dfrac{3+51}{2} = \dfrac{54}{2} = 27$

$x = \dfrac{3-51}{2} = \dfrac{-48}{2} = -24$

But we know that x is positive, so we only accept the solution $x = 27$

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