SOLUTION: i need help solving these quadratic equations using the quadratic formula. 2x^2 - 5x = 3 3x^2 - 2x + 1 = 0

Algebra ->  College  -> Linear Algebra -> SOLUTION: i need help solving these quadratic equations using the quadratic formula. 2x^2 - 5x = 3 3x^2 - 2x + 1 = 0       Log On


   

Question 107040: i need help solving these quadratic equations using the quadratic formula.
2x^2 - 5x = 3
3x^2 - 2x + 1 = 0
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
The quadratic formula starts with the equation:
.
ax%5E2+%2B+bx+%2B+c+=+0
.
in which:
"a" is the multiplier of the term containing x%5E2
"b" is the multiplier of the term containing x
and "c" is the constant.
.
Notice also that the right side of this equation is zero.
.
For an equation of this form, the answer is given by the equation:
.
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
.
Your first equation is 2x%5E2+-+5x+=+3. Notice that this equation is not in the form
required by the quadratic formula because the right side is not zero. Therefore, you need
to make the right side equal to zero, and you can do that by subtracting 3 from both sides
to make the equation become:
.
2x%5E2+-+5x+-3+=+0
.
Now you can compare each term of this equation to the standard quadratic equation.
If you do, you can see that:
.
"a" is +2 because it is the multiplier of the x%5E2 term
"b" is -5 because it is the multiplier of the x term
and "c" is -3 because it is the constant term
.
Now all you need to do is to substitute these values into the equation for x:
.

.
Then all you need to do is simplify the right side of this equation:
.
x+=+%285+%2B-+sqrt%2825+%2B+24%29%29%2F4+=+%285+%2B-+sqrt%2849%29%29%2F4
.
Note that the square root of 49 is +7. Substituting that value into this equation
indicates that you have two answers. The first answer is found by using the + sign in
the numerator to get:
.
x+=+%285%2B7%29%2F4+=+12%2F4+=+3
.
and the second answer is found by using the - sign in the numerator to get:
.
x+=+%285-7%29%2F4+=+-2%2F4+=+-1%2F2
.
So the two answers to this problem are x+=+%2B3 and x+=+-1%2F2
.
This is the same method as you use in the second problem ... you are given the problem:
.
3x%5E2+-+2x+%2B+1+=+0
.
This is already in the standard quadratic form and you can immediately say that:
.
"a" = +3
"b" = -2
and "c" = +1
.
Substitute these values into the equation for the answer x and you get:
.

.
Simplifying the right side:
.
x+=+%282%2B-sqrt%284-12%29%29%2F6=+%282%2B-sqrt%28-8%29%29%2F6
.
This is a little trickier problem than the first one because it contains the square root
of a negative number and therefore you get a complex number as the answer (that is your
answer has both a real part and an imaginary part). If you are familiar with complex numbers,
you may recall that "i" is defined as sqrt%28-1%29 so that i%5E2+=+-1. Therefore,
you can substitute i%5E2 for +-1. When you do, the equation for x becomes:
.

.
Note that sqrt%28i%5E2%29+=+i and sqrt%288%29+=+sqrt%284%2A2%29+=+sqrt%284%29%2Asqrt%282%29+=+2%2Asqrt%282%29.
When you substitute these simplifications into the equation for x it becomes:
.
x+=+%282%2B-i%2A2%2Asqrt%282%29%29%2F6
.
Each of the terms in the numerator contains 2 as a factor. Therefore, you can factor out
this 2 and then divide it by the denominator 6 as follows:
.
x+=+%282%2A%281%2B-i%2Asqrt%282%29%29%29%2F6+=+%281%2B-i%2Asqrt%282%29%29%2F3
.
Then using the + and - signs of the terms in the numerator, you get for your answers:
.
x+=+%281%2Bi%2Asqrt%282%29%29%2F3 and x+=+%281-i%2Asqrt%282%29%29%2F3
.
Hope this clarifies these two problems for you and you can see through all the math manipulations
so that you can understand how to use the quadratic formula.
.