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son4ous [18]
3 years ago
6

Use the quadratic formula to solve the equation. 2x^2-6x+1=0 Enter your answers, in simplified radical form, in the boxes.

Mathematics
2 answers:
Anika [276]3 years ago
5 0

Answer:

(3 (pm) sqrt(7))/2

Step-by-step explanation:

First step: Identify a,b, and c

a=2

b=-6

c=1

Second step: Find b^2-4ac (this is called the discriminant-I will call this D)

(-6)^2-4(2)(1)=36-8=28

Third step: Find the square root of the discriminant aka sqrt(D)

sqrt(28)

Let's see if we can simplify sqrt(28)

Here is there a factor of 28 that is a perfect square? How about 4? Yep!

sqrt(28)=sqrt(4)sqrt(7)=2sqrt(7).

Fourth Step: What is -b? If b=-6 then -b=6.

Fifth step: What is 2a?  2(2)=4

So the formula is

x=(-b (pm) sqrt(D))/(2a)

or

x=(step4 (pm) step 3)/(step 5)

x=(6 (pm) 2sqrt(7))/4

Simplify

x=(3 (pm) sqrt(7))/2

*pm means plus or minus

*sqrt( ) means square root of the number that follows in the ( )

ruslelena [56]3 years ago
5 0

Answer:

Just took it!

Step-by-step explanation:

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What two rational expressions sum to 3x+4/x^2-6x+5?
Lubov Fominskaja [6]

Answer:

\frac{3x+4}{x^2-6x+5}=\frac{-7}{4(x-1)} +\frac{19}{4(x-5)}

Step-by-step explanation:

The given rational expression is:

\frac{3x+4}{x^{2}-6x+5} = \frac{3x+4}{(x-1)(x-5)}

We can use concept of Partial Fractions to solve this problem. Let,

\frac{3x+4}{(x-1)(x-5)}=\frac{A}{x-1} +\frac{B}{x-5}

Multiplying both sides by (x - 1)(x - 5), we get:

3x+4=A(x-5)+B(x-1)

Substituting x = 5, we get:

3(5)+4=A(5-5)+B(5-1)\\\\ 15+4=0+4B\\\\ 19=4B\\\\ B=\frac{19}{4}

Substituting x = 1, we get:

3(1)+4=A(1-5)+B(1-1)\\\\ 7=-4A\\\\ A=-\frac{7}{4}

Substituting the value of A and B, back in the original equation, we get:

\frac{3x+4}{x^2-6x+5}=\frac{-7}{4(x-1)} +\frac{19}{4(x-5)}

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3 years ago
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