# algebra help?? Thanks.?

Ok, so im having alot of trouble with one of my problems for algebra... i have been looking up things to help me on here.. but everyones explanation and steps to get the answer are confusing... heres the equation...

2x^2-3x+1=0...

i know the answer is x= 1 x= 1/2

But i have no idea how people are getting that answer.. could u show me the steps ? Thanks so much :) Peace

Relevance
• Anonymous
8 years ago

2x^2 - 3x + 1 = 0

=> x(1,2) = (-b +- sqrt(b^2 - 4ac)) / 2a

where a = coeff. of x^2, b = coeff. of x and c = constant term

=> x(1,2) = (3 +- sqrt(9 - 8)) / 4

=> x(1,2) = (3 +- 1) / 4

=> x(1,2) = 4/4, 2/4

=> x = 1, 1/2

• I hope this explanation will be clear and easy to understand for you! Here it goes:

Your equation is 2x²-3x+1=0 which is in the form ax²+bx+c=0 where a=2, b=-3 and c=1

When you have an equation with a 'squared term' at the beginning of it, your mind should immediately be thinking 'this is a quadratic equation, so let me try and use the quadratic formula to solve it'.

In case you didn't know, the quadratic formula is:

x=[(-b ± √(b²-4ac)]÷ 2a

Because we already know a=2, b=-3 and c=1 they can all be put straight into the quadratic formula:

x=[(-(-3) ± √(-3²-4(2)(1))]÷ 2(2)

*Note: b is minus 3 and so when put into the quadratic formula, a minus and a minus become a plus. Also, when numbers are together as shown above e.g. 4(2)(1), it means they are being multiplied together so it actually means 4 times 2 times 1= 8 and also 2(2) means 2 times 2 =4

x=[3 ± √(9-8)]÷ 4

x=[3 ± √1]÷ 4

*Note: √1= 1

x=[3 ± 1]÷ 4

There are now two calculations to carry out as there is a plus minus sign, which means that one calculation involves adding the 3 to the 1 and then dividing by 4 and the other involves subtracting the 1 from the 3 then dividing by 4.

Carrying out the calculations one at a time, starting with the plus sign gives:

x=[3 + 1]÷ 4

x=4÷ 4

x=1

Now carrying out the calculation involving the minus sign:

x=[3 - 1]÷ 4

x=2÷ 4

x=1/2

Hope this has helped and if you have any further questions on this explanation feel free to visit the forum at the referenced website below.

Source(s): www.gcsepracticepapers.com
• Factor by grouping method:

multiply first and last term to get 2.

then look for two numbers that add to -3 and MULTIPLY to +2

they are -1 and -2 --> substitute -x and -2x in for -3x in the equation.

2x^2-2x-x+1=0

then split up this equation:

2x^2-2x -x+1

factor: 2x(x-1) - 1(x-1)

(2x-1)(x-1)

2x-1=0 x-1=0

2x=1 x=1

x= 1/2

• Factorize it!

(2*x - 1)*(x - 1) = 0

2*x - 1 = 0, x1 = 1/2

x - 1 = 0, x2 = 1, answer!

• 2x^2-3x+1=0

x=3+ or -/9-4(2) <-square root

___________ <-divided by

2(2)

3+ or -/1 <-square root

_______ <-divided by

2(2)

3+ or -1

________ <-divided by

4

x=4/4 -> 1 or x= 2/4 -> 1/2

My mind went blank as to the name of this method.

• 2x^2-3x+1=0

(2x-1)(x-1)=0

2x-1 = 0

x = 1/2

and x-1=0

x = 1