Here girlfriend go.By the way, when you word her questions, if the number on her worksheet looks like this: 
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Then you write it choose this: 7^2 

x^2 − 5x − 2 = 0.

You are watching: What are the exact solutions of x2 = 5x + 2? x = x = x = x =

a = 1

b = -5

c = -2

x = (5 +- sqrt(25 +8)) / 2

x1 = 5.3723

x2 = -0.37228

Step-by-step explanation:





A quadratic equation

*
,....<1> climate the equipment of this is offered by:

*
....<2>

Given the quadratic equation:

*

We have the right to write this as:

*

On comparing v <1> us have;

a = 1, b = -5 and c = -2

Substitute this in <2> we have

*

*

Simplify:

*

Therefore, the exact solutions the the provided equations are:

*
and also
*



a. X equates to 5 add to or minus the square root of 33, almost everywhere 2

Step-by-step explanation:

x = (5 +- √25+8)/2 = (5 +- √33)/2

sounds prefer x equals 5 add to or minus the square root of 33, almost everywhere 2 come me


A. X equates to 5 add to or minus the square root of thirty-three everywhere 2

Step-by-step explanation:

Let"s relocate all the terms to one side:

*

*

Now, we desire to usage the quadratic formula, which claims that because that a quadratic equation that the kind

*
, the roots can be found with the equation:
*
or
*
.

Here, a = 1, b = -5, and also c = -2, so plug this in:

*

OR

*

Thus, the prize is A.

See more: How Many Covalent Bonds Can Cl Form, Covalent Bond

Hope this helps!


First one:

x = x equals 5 plus or minus the square root of thirty-three anywhere 2

Step-by-step explanation:

x² = 5x + 2

x² - 5x - 2 = 0

Using quadratic formula:

x = <-(-5) +/- sqrt<(-5)² - 4(1)(-2)>/2(1)

x = <5 +/- sqrt(33)>/2


Part 1) x=3

Part 2) x = −1.11 and x = 1.11

Part 3) 105

Part 4) a = −6, b = 9, c = −7

Part 5) x amounts to 5 to add or minus the square source of 33, everywhere 2

Part 6) In the procedure

Part 7)

Part 8) The denominator is 2

Part 9) a = −6, b = −8, c = 12

Step-by-step explanation:

we recognize that

The formula to fix a quadratic equation of the kind is same to

Part 1)

in this problem we have

*

so

*

substitute in the formula

*

*

*

Part 2) in this trouble we have

*

so

*

substitute in the formula

*

*

*

Part 3) as soon as the solution of x2 − 9x − 6 is expressed as 9 to add or minus the square root of r, everywhere 2, what is the value of r?

in this problem we have

*

so

*

substitute in the formula

*

*

therefore

*

Part 4) What space the worths a, b, and c in the following quadratic equation?

−6x2 = −9x + 7

in this difficulty we have

*

*

so

*

Part 5) use the quadratic formula to discover the exact solutions of x2 − 5x − 2 = 0.

In this trouble we have

*

so

*

substitute in the formula

*

*

therefore

x equates to 5 add to or minus the square source of 33, everywhere 2

Part 6) Quadratic Formula proof

we have

Divide both sides by a

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Complete the square

*

*

Rewrite the perfect square trinomial top top the left next of the equation as a binomial squared