Math, asked by poge4056, 1 month ago

When a number is subtracted from its square, the result is 42. Find the number.

Answers

Answered by itzbrainlyBOYZ
2

Answer:

You are wording it wrong. The way you have it worded you end up with imaginary numbers.

You are wording it wrong. The way you have it worded you end up with imaginary numbers.When a number is subtracted from its square the result is 42 !

You are wording it wrong. The way you have it worded you end up with imaginary numbers.When a number is subtracted from its square the result is 42 !x^2-x=42

You are wording it wrong. The way you have it worded you end up with imaginary numbers.When a number is subtracted from its square the result is 42 !x^2-x=42x^2-x-42=0

You are wording it wrong. The way you have it worded you end up with imaginary numbers.When a number is subtracted from its square the result is 42 !x^2-x=42x^2-x-42=0(x-7)(x+6)=0

You are wording it wrong. The way you have it worded you end up with imaginary numbers.When a number is subtracted from its square the result is 42 !x^2-x=42x^2-x-42=0(x-7)(x+6)=0x-7=0

You are wording it wrong. The way you have it worded you end up with imaginary numbers.When a number is subtracted from its square the result is 42 !x^2-x=42x^2-x-42=0(x-7)(x+6)=0x-7=0x=7

You are wording it wrong. The way you have it worded you end up with imaginary numbers.When a number is subtracted from its square the result is 42 !x^2-x=42x^2-x-42=0(x-7)(x+6)=0x-7=0x=77^2-7=49-7=42

You are wording it wrong. The way you have it worded you end up with imaginary numbers.When a number is subtracted from its square the result is 42 !x^2-x=42x^2-x-42=0(x-7)(x+6)=0x-7=0x=77^2-7=49-7=42x+6=0

You are wording it wrong. The way you have it worded you end up with imaginary numbers.When a number is subtracted from its square the result is 42 !x^2-x=42x^2-x-42=0(x-7)(x+6)=0x-7=0x=77^2-7=49-7=42x+6=0x=-6

You are wording it wrong. The way you have it worded you end up with imaginary numbers.When a number is subtracted from its square the result is 42 !x^2-x=42x^2-x-42=0(x-7)(x+6)=0x-7=0x=77^2-7=49-7=42x+6=0x=-6(-6)^2-(-6)=36-(-6)=36+6=42

You are wording it wrong. The way you have it worded you end up with imaginary numbers.When a number is subtracted from its square the result is 42 !x^2-x=42x^2-x-42=0(x-7)(x+6)=0x-7=0x=77^2-7=49-7=42x+6=0x=-6(-6)^2-(-6)=36-(-6)=36+6=427 and -6 are the numbers

Answered by jaydip1118
2

Answer:

There are two numbers which satisfy the criteria

x

=

6

,

x

=

7

Explanation:

Let the number be

=

x

, this number is added to its square

x

2

x

+

x

2

=

42

Now, we solve the equation in order to find

x

x

2

+

x

42

=

0

We first factorise the expression.

We can Split the Middle Term of this expression to factorise it.

x

2

+

x

42

=

x

2

+

7

x

6

x

42

=

x

(

x

+

7

)

6

(

x

+

7

)

=

(

x

6

)

(

x

+

7

)

Equating the factors to zero we get two values for

x

x

=

6

,

x

=

7

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