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4 years ago

Which term completes the product so that it is the difference of squares?

Mathematics

2 answers:

FromTheMoon [43]4 years ago

6 0

Answer:

the answer is 3

Step-by-step explanation:

(-5x-3)(-5x+3)=(5x)^2 - (3)^2

A difference of squares is the difference of two squared terms.

We have

a^2 - b^2

We can factorize the difference of squared terms like this:

a^2-b^2 = (a+b)(a-b)

We have (-5x-3)(-5x+3)

Lets prove it:

(-5x-3)(-5x+3) = (-5x*-5x)+(-5x*3)+(-3*-5x)+(-3*3)

(-5x-3)(-5x+3) = 25x^2+(-15x)+(15x)+(-9)

(-5x-3)(-5x+3) = 25x^2-15x+15x-9

(-5x-3)(-5x+3) = 25x^2-9

(-5x-3)(-5x+3) = (5x)^2 - (3)^2

Ray Of Light [21]4 years ago

5 0

Answer: 3

Step-by-step explanation:

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X^2-8x+16=5 what are the solutions?

svp [43]

Hey!

The first step to solving this equation would be to subtract 5 from both sides of the equation.

Original Equation :
$x^{2}$ - 8x + 16 = 5

New Equation {Added Subtract 5 to Both Sides} :
$x^{2}$ - 8x + 16 - 5 = 5 - 5

Now we solve our new equation.

Old Equation :
$x^{2}$ - 8x + 16 - 5 = 5 - 5

Solution {Old Equation Solved} :
$x^{2}$ - 8x + 11 = 0

The next step would be to solve with the quadratic formula.

Quadratic Equation ( 1 ) :
$x = \frac{-(-8) + \sqrt{(-8)^{2} -4 * 1 * 11 } }{2*1}$

Quadratic Equation ( 1 ) {Solved} :
4 + $\sqrt{5}$

Quadratic Equation ( 2 ) :
$x = \frac{-(-8) - \sqrt{(-8)^{2}-4 *1*11 } }{2*1}$

Quadratic Equation ( 2 ) {Solved} :
4 - $\sqrt{5}$

So, our final solutions for the equation $x^{2}$ - 8x + 16 = 5 are...

x = 4 + $\sqrt{5}$

and 

x = 4 - $\sqrt{5}$

Hope this helps!

- Lindsey Frazier ♥

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