How do I solve this equation? (-4x+12)^1/2 = x

1 answer:

6 0

$(-4x+12)^{1/2}=x\\\sqrt{-4x+12}=x\\\\D:-4x+12\geq0\wedge x\geq0\\D:4x\leq12\wedge x\geq0\\D:x\leq3\wedge x\geq0\\D:x\in\langle0,3\rangle\\\\\sqrt{-4x+12}=x\\-4x+12=x^2\\x^2+4x-12=0\\x^2-2x+6x-12=0\\x(x-2)+6(x-2)=0\\(x+6)(x-2)=0\\x=-6 \vee x=2\\\\-6\not \in D\\\\\boxed{x=2}$

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The domain of both

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Answer:

The domain of function $h(x)$ is set of all real numbers.

Domain: (-∞,∞)

Step-by-step explanation:

Given:

$f(x)=x-6$

$g(x)=x+6$

the domain of both the above functions is all real number.

To find domain of :

$h(x)=f(x)g(x)$

Substituting functions $f(x)$ and $g(x)$ to find $h(x)$

$h(x)=(x-6)(x+6)$

The product can be written as difference of squares. $[a^2-b^2=(a+b)(a-b)]$

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For any quadratic equation the domain is set of all real numbers.

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7 0

4 years ago

Points M and N on the coordinate grid below show the positions of two midfield players of a soccer team:Which statement best des

pogonyaev

N is M reflected across the y-axis; only the signs of the x-coordinates of M and N are different.

<u>Step-by-step explanation:</u>

In question no statements are available , below statements are the required one:

N is M reflected across the x-axis; only the signs of the x-coordinates of M and N are different: N is reflected over y-axis not x-axis so, this option itself is wrong!
N is M reflected across the x-axis; only the signs of the y-coordinates of M and N are different: N-M is reflected over y-axis not x-axis , so this option is also wrong.
N is M reflected across the y-axis; only the signs of the x-coordinates of M and N are different: N-M are reflected over y-axis correct and sign of only x co-ordinate is changed as Co-ordinates are $M(-2,-2.5)$ and $N(2,-2.5)$ .Hence, this statement is correct!
N is M reflected across the y-axis; only the signs of the y-coordinates of M and N are different. N-M is reflected over y-axis , and sign of only x co-ordinate is changed not y . False statement.