One solution of x^2-64=0 is 8. What is the other solution?

Mathematics

1 answer:

matrenka [14]3 years ago

8 0

Answer:

-8

Step-by-step explanation:

Multiplying two negatives (or squaring them) cancels the negative, therefore, (-8)^2=64

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Factor each.<br><br><br> Please help!

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Answer:

1) The factors of $x^2-3x+2=0$ are $\mathbf{(x-1)(x-2)=0}$

Option C is correct.

3) The factors of $x^2+2x-8=0$ are $\mathbf{(x-2)(x+4)=0}$

Option B is correct.

Step-by-step explanation:

1) Factor : $x^2-3x+2=0$

For factoring we need to break the middle term, such that there sum is equal to middle term of expression and product is equal to product of first and last term.

The middle term : -3x

We can break them as (-2x)( -x)

Solving:

$x^2-3x+2=0\\x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0$

So, factors of $x^2-3x+2=0$ are $\mathbf{(x-1)(x-2)=0}$

Option C is correct.

3) Factor: $x^2+2x-8=0$

For factoring we need to break the middle term, such that there sum is equal to middle term of expression and product is equal to product of first and last term.

The middle term : 2x

We can break them as (4x)( -2x)

Solving

$x^2+2x-8=0\\x^2+4x-2x-8=0\\x(x+4)-2(x+4)=0\\(x-2)(x+4)=0$