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 alsoa. 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.

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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

Complete the square

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