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3 years ago

Complete the equation so that it has solutions of -5 and 7 2+

Mathematics

1 answer:

lorasvet [3.4K]3 years ago

7 0

Answer:

x^2 -2x -35 = 0

Step-by-step explanation:

Lets's solve backwards

x = -5

Add 5 to both sides

x + 5 = 5 - 5

x + 5 = 0

and

x = 7

Subtract 7 from both sides

x - 7 = 7 - 7

x - 7 = 0

Finding the product of both linear functions

(x + 5)(x - 7) =0

x^2 -7x + 5x -35 =0

x^2 -2x -35 = 0

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What are the solution(s) to the quadratic equation 40 - x2 = 0?

Zanzabum

Answer:

$x=\pm2\sqrt{10}$

Step-by-step explanation:

You want to solve ...

$40 -x^2=0\\\\40=x^2\qquad\text{add $x^2$}\\\\\pm\sqrt{40}=x\qquad\text{take the square root}\\\\\boxed{x=\pm2\sqrt{10}}\qquad\text{simplify}$

5 0

3 years ago

Please help i need this

Svet_ta [14]

Slope Formula - (y² - y¹) / (x² - x¹)

(1, 7)

x¹ = 1

y¹ = 7

(-4, -8)

x² = -4

y² = -8

-8 - 7 = -15

-4 - 1 = -5

Simplify - -15/-5

-15/5 = -3

Your answer is -3.

Hope this Helps!!

5 0

3 years ago

Simplify, state all restrictions.

Kipish [7]

The simplified expression is $\frac{1 - x - y}{x + y}$ and the restriction is $y \ne -x$

<h3>How to simplify the expression?</h3>

The expression is given as:

$\frac{x - y}{4x^2 - 8xy + 3y^2} \div \frac{2x + y}{2x - 3y} \times \frac{4x^2 - y^2}{x^2 - y^2} -1$

Express x^2 - y^2 as (x + y)(x - y) and factorize other expressions

$\frac{x - y}{(2x - y)(2x - 3y)} \div \frac{2x + y}{2x - 3y} \times \frac{4x^2 - y^2}{(x - y)(x + y)} -1$

Rewrite the expression as products

$\frac{x - y}{(2x - y)(2x - 3y)} \times \frac{2x - 3y}{2x + y} \times \frac{4x^2 - y^2}{(x - y)(x + y)} -1$

Cancel out the common factors

$\frac{1}{(2x - y)} \times \frac{1}{2x + y} \times \frac{4x^2 - y^2}{(x + y)} -1$

Express 4x^2 - y^2 as (2x - y)(2x + y)

$\frac{1}{(2x - y)} \times \frac{1}{2x + y} \times \frac{(2x - y)(2x + y)}{(x + y)} -1$

Cancel out the common factors

$\frac{1}{x + y} -1$

Take the LCM

$\frac{1 - x - y}{x + y}$

Hence, the simplified expression is $\frac{1 - x - y}{x + y}$ and the restriction is $y \ne -x$