Solve by factoring: 3x^2 + 5x - 2 = 0. Solve by factoring. 1. x^2 + 5x + 6 = 0 2. 2x^2 - x - 3 = 0 3. 3x^2 + 5x - 2 = 0
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1. x^2 + 5x + 6 = 0
Begin with this set up:
(x + __)(x + __) = 0
The missing numbers must have a product of 6 and a sum of 5.
2*3=6 2+3=5
Therefore...
(x + 2)(x + 3) = 0
Now take each binomial and set it equal to 0. Solve for x.
x + 2 = 0 x = -2
x + 3 = 0 x = -3
Solution set: {-2, -3}
2. 2x^2 - x - 3 = 0
When there are two subtraction signs in the trinomial, the set up is:
(2x + __)(x - __) = 0 or (2x - __)(x + __) = 0
Now we need two numbers whose product is 3. Since 3 is prime, the only options are 1 and 3. Try substituting 1 and 3 into the set ups above, use FOIL to see which set up works.
(2x + 1)(x - 3) = 2x^2 - 6x + 1x - 3 = 2x^2 - 5x - 3 NO
(2x + 3)(x - 1) = 2x^2 - 2x + 3x - 3 = 2x^2 + 1x - 3 NO
(2x - 1)(x + 3) = 2x^2 + 6x - 1x - 3 = 2x^2 + 5x - 3 NO
(2x - 3)(x + 1) = 2x^2 + 2x - 3x - 3 = 2x^2 - 1x - 3 YES
So now we know the trinomial can be factored as...
(2x - 3)(x + 1) = 0
Again, set each binomial equal to 0 and solve for x.
2x - 3 = 0 x = 1.5
x + 1 = 0 x = -1
Solution set: {1.5, -1}
3. 3x^2 + 5x - 2 = 0
This one is solved similarly to #2. Here is the set up:
(3x + __)(x - __) = 0 or (3x - __)(x + __) = 0
Again, since 2 is prime, the only options for the blanks are 1 and 2. Try each combination, use FOIL to see which one works.
(3x + 1)(x - 2) = 3x^2 - 6x + 1x - 2 = 3x^2 - 5x - 2 NO
(3x + 2)(x - 1) = 3x^2 - 3x + 2x - 2 = 3x^2 + 1x - 2 NO
(3x - 1)(x + 2) = 3x^2 + 6x - 1x - 2 = 3x^2 + 5x - 2 YES
Therefore...
(3x - 1)(x + 2) = 0
Set each binomial equal to 0 and solve for x.
3x - 1 = 0 x = 1/3
x + 2 = 0 x = -2
Solution set: {1/3, -2}
Remember, you can always check these answers by graphing the equations. The x-intercepts should equal the solution set.
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You are allowed to ask only one question at a time. I am solving one of the problems, you can apply the same method for the others.
To solve 3x^2 + 5x - 2 = 0, we express 5 in terms of two numbers that add up to 5 and the product of the numbers is 3*-2 = -6. The numbers that satisfy this are 6 and -1.
3x^2 + 5x - 2 = 0
=> 3x^2 + 6x - x - 2 = 0
=> 3x(x + 2) - 1(x + 2) = 0
=> (3x - 1)(x + 2) = 0
3x - 1 = 0
=> x = 1/3
x + 2 = 0
=> x = -2
The solution of 3x^2 + 5x - 2 = 0 is x = 1/3 and x = -2
Going to do #2
2x^2 - x - 3 = 0
a b c
multiply a by c
2 x -3 = -6
find factors of -6 that minus to b (-1) which would be -3 and 2
plug those numbers in as b
2x^2 + 2x - 3x - 3
group
(2x^2 + 2x)( - 3x - 3)
factor out
2x ( x +1) -3 (x +1)
group the numbers outside together:
(2x - 3 ) (x + 1)
that's the answer but you can go further to find the solutions by setting the parentheses = 0
2x - 3 =0
2x = 3
x = 3 / 2 or 1.5
x + 1 = 0
x = 0
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