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3 years ago

ASAP PLS!! i attached a link with a screenshot of the question. please give step by step!!!!! will mark brainliest :)

Mathematics

1 answer:

emmainna [20.7K]3 years ago

7 0

Answer:

I found two different solutions. Hope one of them help!

1. x = -1/3 = -0.333

2. x = 5/2 = 2.500

Step-by-step explanation:

13 ± √ 289

x = ——————

Can √ 289 be simplified ?

Yes! The prime factorization of 289 is

17•17

To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).

√ 289 = √ 17•17 =

± 17 • √ 1 =

± 17

So now we are looking at:

x = ( 13 ± 17) / 12

Two real solutions:

x =(13+√289)/12=(13+17)/12= 2.500

x =(13-√289)/12=(13-17)/12= -0.333

Two solutions were found :

x = -1/3 = -0.333

x = 5/2 = 2.500

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Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

kogti [31]

Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

$x = a$ (2)

$(b^{2}-1) = 2\cdot x\cdot y$ (3)

$y = c$ (4)

By (2) and (4) in (3), we have the following expression:

$(b^{2} - 1) = 2\cdot a \cdot c$

$b^{2} = 1 + 2\cdot a \cdot c$

$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.