# Solve the following using the quadratic formula. Simplify and give exact answers only. State imaginary answers in terms of “i”. a)3x2 + x –2 = 0b) x2 – 5x + 7 = 0

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The quadratic formula is x= -b +- sqr root(bsquared - 4ac) all divided by 2a.

In part a) a=3, b=1 and c=-2 so, [-1 (+-) sqr root(1squared - 4*3*(-2))]/(2*3) gives x= [-1 (+-) sqr root (1+24)]/6

Gives x= [-1 (+-)sqr root (25)]/6. sqr root 25 = 5, so you have [-1(+-)5]/6. Complete [-1+5]/6 and [-1-5]/6 to get x=4/6 which reduces to 2/3 and x=-6/6 which reduces to -1.

So answers to part a) are 2/3 and -1

Part b) will be completed similarly with a=1, b=-5 and c=7.

x=[5(+-)sqr root((-5)squared - (4*1*7)0]/(2*1)

Which reduces to [5(+-)sqr root(25 - 28)]/2. 25 - 28 = (-3) so, x=[5(+-)sqr root (-3)]/2. Sqr root of (-3) is i*sqr root of 3.

Therefore, x=[5(+-) i sqr root (3)]/2

a)3x^2 + x –2 = 0

Applying quadratic formula:

`x= {-b+-sqrt(b^2-4ac)} / (2a)`

` `Where

a=3

b=1

c=-2

Input the values

`x={-1+-sqrt(1^2-4*3*-2)} /(2*3)`

`x= {-1+-sqrt(1+24)} / 6`

`x= (-1+-5)/ 6`

`x= 4/6`

`x= -6/6`

`Solution Set= {2/3,-1}`

b) x2 – 5x + 7 = 0

Where

a=1

b=-5

c=7

Input the values

`x= {-(-5)+-sqrt(-5^2-4*1*7)} / (2*1)`

`x= {5+-sqrt(25-28)} /2`

`x= {5+-sqrt(-3)} /2`

`x= {5+-i sqrt(3)} /2`