Question

find x in equation

${x}^{2} + \sqrt{625} = 0$
​

Answer 1

Answer:

5 I okk please mark me as brianlist and thanks my all answers please.

Answer 2

Answer:

The answer is 5

The first term is, x2 its coefficient is 1 .

The middle term is, -50x its coefficient is -50 .

The last term, "the constant", is +625

Step-1 : Multiply the coefficient of the first term by the constant 1 • 625 = 625

Step-2 : Find two factors of 625 whose sum equals the coefficient of the middle term, which is -50 .

-625 + -1 = -626

-125 + -5 = -130

-25 + -25 = -50 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -25 and -25

x2 - 25x - 25x - 625

Step-4 : Add up the first 2 terms, pulling out like factors :

x • (x-25)

Add up the last 2 terms, pulling out common factors :

25 • (x-25)

Step-5 : Add up the four terms of step 4 :

(x-25) • (x-25)

Which is the desired factorization

Multiplying Exponential Expressions:

1.2 Multiply (x-25) by (x-25)

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is (x-25) and the exponents are :

1 , as (x-25) is the same number as (x-25)1

and 1 , as (x-25) is the same number as (x-25)1

The product is therefore, (x-25)(1+1) = (x-25)2

Equation at the end of step

1

:

(x - 25)2 = 0