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Dmitrij [34]
3 years ago
5

X^2-25=0 Using factoring method

Mathematics
2 answers:
Tema [17]3 years ago
6 0
Hello.

Move <span><span>−25</span><span>-25</span></span> to the right side of the equation by subtracting <span><span>−25</span><span>-25</span></span> from both sides of the equation.<span><span><span>x2</span>=25</span><span><span>x2</span>=25

</span></span>Take the square root of both sides of the equation to eliminate the exponent on the left side.
<span><span>
</span></span>The complete solution is the result of both the positive and negative portions of the solution.

Simplify the right side of the equation.

Rewrite <span>2525</span> as <span><span>52</span><span>52</span></span>.

Pull terms out from under the radical, assuming positive real numbers.The complete solution is the result of both the positive and negative portions of the solution.
<span><span>
</span></span>The complete solution is the result of both the positive and negative portions of the solution.

<span>Answer: x=5,−<span>5

Have a nice day</span></span>
Alexxandr [17]3 years ago
5 0
x^2  - 25 = 0

That's the difference of two squares, which has a general factorization 

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

Applying it to our case

0 = x^2 - 25 = (x+5)(x-5)

Two factors multiply to zero; either must be zero.

x=-5 \textrm{ or } x=5



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Without plotting points, let M=(-2,-1), N=(3,1), M'= (0,2), and N'=(5, 4). Without using the distanceformula, show that segments
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Given:

M=(x1, y1)=(-2,-1),

N=(x2, y2)=(3,1),

M'=(x3, y3)= (0,2),

N'=(x4, y4)=(5, 4).

We can prove MN and M'N' have the same length by proving that the points form the vertices of a parallelogram.

For a parallelogram, opposite sides are equal

If we prove that the quadrilateral MNN'M' forms a parallellogram, then MN and M'N' will be the oppposite sides. So, we can prove that MN=M'N'.

To prove MNN'M' is a parallelogram, we have to first prove that two pairs of opposite sides are parallel,

Slope of MN= Slope of M'N'.

Slope of MM'=NN'.

\begin{gathered} \text{Slope of MN=}\frac{y2-y1}{x2-x1} \\ =\frac{1-(-1)}{3-(-2)} \\ =\frac{2}{5} \\ \text{Slope of M'N'=}\frac{y4-y3}{x4-x3} \\ =\frac{4-2}{5-0} \\ =\frac{2}{5} \end{gathered}

Hence, slope of MN=Slope of M'N' and therefore, MN parallel to M'N'

\begin{gathered} \text{Slope of MM'=}\frac{y3-y1}{x3-x1} \\ =\frac{4-(-1)}{5-(-2)} \\ =\frac{3}{2} \\ \text{Slope of NN'=}\frac{y4-y2}{x4-x2} \\ =\frac{4-1}{5-3} \\ =\frac{3}{2} \end{gathered}

Hence, slope of MM'=Slope of NN' nd therefore, MM' parallel to NN'.

Since both pairs of opposite sides of MNN'M' are parallel, MM'N'N is a parallelogram.

Since the opposite sides are of equal length in a parallelogram, it is proved that segments MN and M'N' have the same length.

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Cerrena [4.2K]

Answer:

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Step-by-step explanation:

\sqrt{57} =\sqrt{3 * 19}

Since this cannot be further simplified, we will calculate the square root of 57 with our calculators.

We find that the square root of 57 is 7.549834435, and since the tenths place is a 5, we will round up to the next whole number. So, the point on the number line that best represents the square root of 57 is 8.

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