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Sedaia [141]
3 years ago
9

Solve the equation -x^2+50=25

Mathematics
2 answers:
DiKsa [7]3 years ago
5 0

Answer:

Two solutions were found :

x = 5

x = -5

Step-by-step explanation:

Step  1  :

Trying to factor as a Difference of Squares :

1.1      Factoring:  25-x2

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =

        A2 - AB + BA - B2 =

        A2 - AB + AB - B2 =

        A2 - B2

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check :  25  is the square of  5

Check :  x2  is the square of  x1

Factorization is :       (5 + x)  •  (5 - x)

Equation at the end of step  1  :

 (x + 5) • (5 - x)  = 0

Step  2  :

Theory - Roots of a product :

2.1    A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

2.2      Solve  :    x+5 = 0

Subtract  5  from both sides of the equation :

                     x = -5

Solving a Single Variable Equation :

2.3      Solve  :    -x+5 = 0

Subtract  5  from both sides of the equation :

                     -x = -5

Multiply both sides of the equation by (-1) :  x = 5

Two solutions were found :

x = 5

x = -5

Processing ends successfully

plz mark me as brainliest if this helped :)

Tema [17]3 years ago
3 0

Answer:

x = ±5

Step-by-step explanation:

-x²+50=25   (subtract 50 from both sides)

-x² = 25 - 50

-x² = -25  (multiply both sides by -1)

x² = 25  (take square root of both sides)

x = ±√25   (recall that 5 x 5 = 25)

x = ±5

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Answer:

  (x, y) = (1/2, -1)

Step-by-step explanation:

Subtracting twice the first equation from the second gives ...

  (2/x +1/y) -2(1/x -5/y) = (3) -2(7)

  11/y = -11 . . . . simplify

  y = -1 . . . . . . . multiply by y/-11

Using the second equation, we can find x:

  2/x +1/-1 = 3

  2/x = 4 . . . . . . . add 1

  x = 1/2 . . . . . . . multiply by x/4

The solution is (x, y) = (1/2, -1).

_____

<em>Additional comment</em>

If you clear fractions by multiplying each equation by xy, the problem becomes one of solving simultaneous 2nd-degree equations. It is much easier to consider this a system of linear equations, where the variable is 1/x or 1/y. Solving for the values of those gives you the values of x and y.

A graph of the original equations gives you an extraneous solution of (x, y) = (0, 0) along with the real solution (x, y) = (0.5, -1).

6 0
2 years ago
Find two equivalent ratios for each ratio.
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Choose the best answer:
Oksi-84 [34.3K]

Answer:

The length of the hypotenuse is 2 square root of 13 ⇒ c

Step-by-step explanation:

The rule of the area of the right triangle is A = \frac{1}{2} × leg1 × leg2, where

leg1 and leg2 are the sides of the right angle

∵ The area of a right triangle is 12 in²

∵ The ratio of the length of its legs is 2: 3

→ Let leg1 = 2x and leg2 = 3x

∵ leg1 = 2x and leg2 = 3x

→ Substitute them in the rule of the area above

∴ 12 = \frac{1}{2} × 2x × 3x

∵ 2x × 3x = 6x²

∴ 12 =  \frac{1}{2} × 6x²

∴ 12 = 3x²

→ Divide both sides by 3 to find x²

∴ 4 = x²

→ Take √ for both sides

∴ x = 2

→ Substitute x in the expressions of leg1 and leg2 to find them

∴ leg1 = 2(2) = 4 inches

∴ leg2 = 3(2) = 6 inches

∵ hypotenuse = \sqrt{(leg1)^{2}+(leg2)^{2}}

∴ hypotenuse = \sqrt{(4)^{2}+(6)^{2}}=\sqrt{16+36}=\sqrt{52}

∵ The simplest form of \sqrt{52} = 2\sqrt{13}

∴ The length of the hypotenuse = 2\sqrt{13} inches

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Answer:

No

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