6p² + p = 1
If someone would show me how to work this problem, step by step, I'd appreciate it greatly! :D
If someone would show me how to work this problem, step by step, I'd appreciate it greatly! :D
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First method, use the quadratic formula. In the standard form, we have
6p² + p - 1 = 0
so, the variable is p, and the coefficients will be a = 6, b =1, and c = -1. Thus,
p = [ -b +/- sqrt(b^2-4ac) ]/(2a) = [-1 +/- sqrt( 1^2 - 4*6*(-1)) ]/12 = [-1 +/- 5]/12
now, the +/- provides both solutions, and evaluating them separately we get
p = { (-1+5)/12, (-1-5)/12 } = { 1/3, -1/2 }.
Method 2: Factor
First we divide by 6
p² + p/6 - 1/6 = 0
and seek a factorization
(p-p1)*(p-p2) = 0
which after foiling gives
p² - (p1+p2)*p + p1*p2 = 0
The first expression tells us p = p1 and p = p2 are the solutions, and the second tells us that
p1 + p2 = -1/6
p1 * p2 = -1/6
We then look for the two numbers p1 and p2 whose sum is -1/6, and whose product is -1/6. The candidates will be 1/2, -1/2, 1/3, -1/3, 1, -1, 1/6 and -1/6; otherwise, the fractions will need to involve irrational numbers with square roots.
After some guesswork, it turns out that the correct pair will be -1/2 and 1/3, since
-1/2 + 1/3 = -3/6 + 2/6 = -1/6
and
(-1/2) * 1/3 = -1/6.
Thus, p1 = -1/2 and p2 = 1/3, so that
p = { 1/3, -1/2 }.
6p² + p - 1 = 0
so, the variable is p, and the coefficients will be a = 6, b =1, and c = -1. Thus,
p = [ -b +/- sqrt(b^2-4ac) ]/(2a) = [-1 +/- sqrt( 1^2 - 4*6*(-1)) ]/12 = [-1 +/- 5]/12
now, the +/- provides both solutions, and evaluating them separately we get
p = { (-1+5)/12, (-1-5)/12 } = { 1/3, -1/2 }.
Method 2: Factor
First we divide by 6
p² + p/6 - 1/6 = 0
and seek a factorization
(p-p1)*(p-p2) = 0
which after foiling gives
p² - (p1+p2)*p + p1*p2 = 0
The first expression tells us p = p1 and p = p2 are the solutions, and the second tells us that
p1 + p2 = -1/6
p1 * p2 = -1/6
We then look for the two numbers p1 and p2 whose sum is -1/6, and whose product is -1/6. The candidates will be 1/2, -1/2, 1/3, -1/3, 1, -1, 1/6 and -1/6; otherwise, the fractions will need to involve irrational numbers with square roots.
After some guesswork, it turns out that the correct pair will be -1/2 and 1/3, since
-1/2 + 1/3 = -3/6 + 2/6 = -1/6
and
(-1/2) * 1/3 = -1/6.
Thus, p1 = -1/2 and p2 = 1/3, so that
p = { 1/3, -1/2 }.
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6p² + p = 1
Step 1, Subtract 1 to both sides of the equal sign. Making the equation equal to zero.
6p² + p - 1 = 1 - 1
6p² + p - 1 = 0
Step 2, Perform the FOIL method.
(3p - 1)(2p + 1) = 0
Step 3, Set each group equal to zero and solve for p.
3p - 1 = 0
3p - 1 + 1 = 0 + 1
3p = 1
3p / 3 = 1 / 3
p = 1/3
2p + 1 = 0
2p + 1 - 1 = 0 - 1
2p = - 1
2p / 2 = - 1 / 2
p = - 1/2
The answers are p = 1/3 or - 1/2
Step 1, Subtract 1 to both sides of the equal sign. Making the equation equal to zero.
6p² + p - 1 = 1 - 1
6p² + p - 1 = 0
Step 2, Perform the FOIL method.
(3p - 1)(2p + 1) = 0
Step 3, Set each group equal to zero and solve for p.
3p - 1 = 0
3p - 1 + 1 = 0 + 1
3p = 1
3p / 3 = 1 / 3
p = 1/3
2p + 1 = 0
2p + 1 - 1 = 0 - 1
2p = - 1
2p / 2 = - 1 / 2
p = - 1/2
The answers are p = 1/3 or - 1/2
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6p² + p = 1
6p² + p -1= 0
6p² +3p-2p-1
3p(2p+1)-1(2p+1)
(3p-1)(2p+1)
6p² + p -1= 0
6p² +3p-2p-1
3p(2p+1)-1(2p+1)
(3p-1)(2p+1)