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OlgaM077 [116]
3 years ago
14

Please help!

Mathematics
1 answer:
crimeas [40]3 years ago
4 0

Answer:

The zeros are:

x =4, x=2, x = 5

  • The function has three distinct real zeros.

Hence, option (B) is true.

Step-by-step explanation:

Given the expression

h\left(x\right)=\left(x-4\right)^2\left(x^2-7x+\:10\right)

Let us determine the zeros of the function by putting h(x) = 0 and solving the expression

0=\left(x-4\right)^2\left(x^2-7x+10\right)

switch sides

\left(x-4\right)^2\left(x^2-7x+10\right)=0

as

x^2-7x+\:10=\left(x-2\right)\left(x-5\right)

so

\left(x-4\right)^2\left(x-2\right)\left(x-5\right)=0

Using the zero factor principle

  • \mathrm{If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

so

x-4=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

x =4, x=2, x = 5

Thus, the zeros are:

x =4, x=2, x = 5

It is clear that there are three zeros and all the zeros are distinct real numbers.

Therefore,

  • The function has three distinct real zeros.

Hence, option (B) is true.

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The problem:

What are x values that satisfy:

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Step-by-step explanation:

No number can be substituted into x+4=x+8 to make it true.

There is no number that you can find such that when you add 4 to it will give you the same as adding 8 to it.

Also if you subtract x on both sides you obtain the equation 4=8.

4=8 is not true so x+4=x+8 is never true for any x.

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