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3 years ago

If 4x^2-100=0, the roots are

1 answer:

8 0

This is a difference of 2 perfect squares where you can factor x^2-y^2=(x-y)(x+y) so

also if it is equal to zero and you factor, then each factor is assumed set to zero exg
xy=0 then assume x and/or y=0 so

4x^2-100
(2x)^2-(10)^2=(2x-10)(2x+10)=0
so the roots are (2x-10) and (2x+10)
(solution is 5 and -5)

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3 years ago

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vampirchik [111]

Answer:

Step-by-step explanation:

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3 years ago

If k stands for an integer, then is it possible for k2 + k to stand for an odd integer? Be prepared to justify your answer.

BartSMP [9]

Answer:

k^2 + k never stands for an odd integer

Step-by-step explanation:

Let us consider either case with which k stands for an odd or even integer;

Case 1: k is an odd integer

For integer a, k = 2a + 1

So, k + 1 = 2a + 2 = 2( a + 1 ) = 2b for integer b

k^2 + k = k ( k + 1 ) = k ( 2b ) = 2kb = 2c for integer c,

Therefore, if k is an odd integer, then k^2 + k is an even integer ;

Case 2: k is an even integer

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So, k + 1 = 2a + 1

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Therefore, if k is an even integer, then k^2 + k is an even integer;

This would make k^2 + k never stand for an odd integer

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3 years ago

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Step-by-step explanation:

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