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Setler79 [48]
3 years ago
10

Solve for t(advanced rational equations)

Mathematics
2 answers:
Mariulka [41]3 years ago
5 0
<u>t² -7 + 12t⁻² = 0</u>

Multiply each side by t² :

t⁴ -7t² + 12 = 0

This is a quadratic equation in the variable ' t² '.
For just a moment, to avoid confusion, let U = t².
Then the equation is

U² - 7U + 12 = 0
whence
(U - 4) (U - 3) = 0
and
U = 4   and   U = 3 .

Now we can go back to ' t² ' in place of U :

t² = 4
<u>t = +2</u>  and  <u>t = -2</u>

t² = 3
<u>t = +√3</u>  and <u> t = -√3</u>

                                       
Hoochie [10]3 years ago
5 0
t^2-7+12t^{-2}=0\\&#10;t^4-7t^2+12=0\\&#10;t^4-3t^2-4t^2+12=0\\&#10;t^2(t^2-3)-4(t^2-3)=0\\&#10;(t^2-4)(t^2-3)=0\\&#10;(t-2)(t+2)(t^2-3)=0\\&#10;t=2 \vee t=-2 \vee t=-\sqrt{3}\vee t=\sqrt3&#10;
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Use the identity (x2+y2)2=(x2−y2)2+(2xy)2 to determine the sum of the squares of two numbers if the difference of the squares of
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Answer:

The sum of the squares of two numbers whose difference of the squares of the numbers is 5 and the product of the numbers is 6 is <u>169</u>

Step-by-step explanation:

Given :  the difference of the squares of the numbers is 5 and the product of the numbers is 6.

We have to find the sum of the squares of two numbers whose difference and product is given using given identity,

(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2

Since, given the difference of the squares of the numbers is 5 that is (x^2-y^2)^2=5

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Using identity, we have,

(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2

Substitute, we have,

(x^2+y^2)^2=(5)^2+(2(6))^2

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(x^2+y^2)^2=25+144

(x^2+y^2)^2=169

Thus, the sum of the squares of two numbers whose difference of the squares of the numbers is 5 and the product of the numbers is 6 is 169

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