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krek1111 [17]
3 years ago
8

Choose all the quadratic pairs that are equivalent. 1 2 (x + 4)2 = 2 x2 + 8x + 15 = 0 (x − 5)2 = 1 x2 − 10x + 26 = 0 3(x − 1)2 +

5 = 0 3x2 − 6x + 8 =
Mathematics
1 answer:
Artemon [7]3 years ago
3 0

Answer:

Only equation 1 and 2 are equal.

Step-by-step explanation:

2 (x + 4)2 = 2

2( x² + 8x+ 16) = 2               Applying the square formula

2x² + 16x+ 32 = 2

2x² + 16x+ 32 -2= 0

2x² + 16x+ 30 = 0

2( x² + 8x+ 15)= 0           Taking 2 as common

x2 + 8x + 15 = 0------------eq 1

x2 + 8x + 15 = 0-------------eq 2

(x − 5)2 = 1

x²-10x+25= 1             Applying the square formula

x²-10x+25- 1= 0

x²-10x+24= 0-------------eq 3

x2 − 10x + 26 = 0 -------------eq 4

3(x − 1)2 + 5 = 0

3( x²-2x+1)+5= 0               Applying the square formula

3x²-6x+3+5= 0

3x²-6x+ 8= 0-------------eq 5

3x2 − 6x + 8 =1

3x2 − 6x + 8 -1=0

3x2 − 6x + 7 =0-------------eq 6

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2 years ago
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
fomenos

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

And the product of the numbers is 6 that is xy=6

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

Simplify, we have,

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(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

5 0
3 years ago
Read 2 more answers
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