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3 years ago

How can I get an equation that fits this

Mathematics

1 answer:

aleksandr82 [10.1K]3 years ago

5 0

Answer:

Step-by-step explanation:

non real roots come in pairs. so an example of a non real root would be y=ax + bi and y = ax - bi.

the fact that it is rational with a multiplicity of 2 means that their is a root that is raised to the power of 2. So we now have y = (x +a)^2 or something like it.

The leading coefficient is even. That means y = 2c * (x + 2)^2 (ax + bi)(ax - bi)

The fact that the two variables (x and y) both go to minus infinity means that c<0. c is minus).

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3x-(2x-1)=7x-(3*5x)+(-x+24)

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Answer:

$\boxed{\boxed{\sf x=\frac{23}{10}}\: \sf or \:\boxed{x=2.3}}$

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$\boxed{\sf Step\: By\:Step:- }$

$\sf 3x-\left(2x-1\right)=7x-\left(3\times \:5x\right)+\left(-x+24\right)$

Remove the parentheses:

$\to\sf 3x-\left(2x-1\right)=7x-3\times \:5x-x+24$

Combine like terms:

$\sf ^*7x-x=6x$

$\to\sf 3x-\left(2x-1\right)=6x-3\times \:5x+24$

Multiply 3 and 5x = 15x:-

$\to\sf 3x-\left(2x-1\right)=6x-15x+24$

Combine like terms:

$\sf ^*6x-15x=-9x$

$\to\sf 3x-\left(2x-1\right)=-9x+24$

Expand: 3x-(2x-1)= x+1

$\to\sf x+1=-9x+24$

Subtract 1 from both sides:

$\to\sf x+1-1=-9x+24-1$

$\to\sf x=-9x+23$

Add 9x to both sides:

$\to\sf x+9x=-9x+23+9x$

$\to\sf 10x=23$

Divide both sides by 10:

$\to\sf \cfrac{10x}{10}=\cfrac{23}{10}$

$\to\sf x=\cfrac{23}{10}$

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