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son4ous [18]
3 years ago
7

Find the real zeros x4-11x2+18=0

Mathematics
1 answer:
masya89 [10]3 years ago
6 0

Answer:

x=√2, x=-√2, x= 3 and x=-3

Step-by-step explanation:

We need to solve the equation x^4 - 11x^2+18=0

We can replace x^4 = u^2 and x^2 = u

So, the equation will become

u^2 -11u+18 = 0

Factorizing the above equation:

u^2 -9u-2u+18 =0

u(u-9)-2(u-9)=0

(u-2)(u-9)=0

u=2, u=9

As, u = x^2, Putting back the value:

x^2 =2 , x^2 =9

taking square roots:

√x^2 =√2 ,√x^2=√9

x=±√2 , x = ±3

so, x=√2, x=-√2, x= 3 and x=-3

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Seven less than one-fourth of a number is 2. What is the number?
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Let the number be represented by x

Therefore, seven less than one-fourth of a number will be:

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Learn more about equation on:

brainly.com/question/13802812

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