Question 28502: 1) Using the quadratic equation 2/3^2+X=1/2, perform the following Found 3 solutions by LilSkittleMd, sdmmadam@yahoo.com, LB1973:Answer by LilSkittleMd(119) (Show Source):
You can put this solution on YOUR website! b)Solve by completing the square.
Show work in this space.
c)Solve by using the quadratic formula.
Show work in this space.
Solve by factoring:
(x-2) (x-1)
Set each equal to 0
x-2=0 x-1=0
Solve each equation
x=2 x=1
Solve by completing the square:
Subtract 2 from both sides to get the x's on one side of the equation, and any numbers on the other side.
Take the middle term, divide by 2, and square it.
Then add that number to both sides of the equation. Now you have:
Now take x, the sign after and the answer of the middle term divided by 2, put them in parenthesis and square it.
Now use the square root method to solve the problem. Square root both sides of the equation to get rid of the square. This gives you:
Simplify. This gives you:
Now add to both sides
x=2 x=1
You can put this solution on YOUR website! 1) Using the quadratic equation x^2 - 3x + 2 = 0, perform the following tasks:
a) Solve by factoring.
Answer:
Show work in this space.
b) Solve by completing the square.
Show work in this space.
c) Solve by using the quadratic formula.
Show work in this space.
Exactly how much to present in an examination (without books!)
Factoring Method
x^2 - 3x + 2 = 0-----(*)
x^2+[(-2x)+(-x)]+2=0
(x^2-2x)-x+2=0 (by additive associativity)
x(x-2)-1(x-2)=0
xp-p=0 where p =(x-2)
p(x-1)=0
which is (x-2)(x-1)=0
(x-2) = 0 implies x = 2
(x-1) = 0 implies x = 1
Answer: x = 2 and x=1
Completing the square method
x^2 - 3x + 2 = 0-----(*)
x^2 - 3x = -2
(x)^2-2X(3/2)Xx = -2
[(x)^2-2X(3/2)Xx +(9/4) = -2 +(9/4)
(X-3/2)^2 = (-8+9)/4
(x-3/2)^2 = 1/4
(x-3/2) = (+ or minus)(1/2)
x = +3/2+ or minus(1/2)
That is
x = 3/2+1/2 = 2 or x= 3/2-1/2 = 1
Answer: x = 2 and x = 1
Formula method
x^2 - 3x + 2 = 0-----(*)is of the form (a)x^2 +bx+c = 0
with a = 1, b =(-3) and c= 2
By formula x = {(-b)+or minus[sqrt(b^2-4ac)]}/(2a)}
x= [3+ or minus 1]/2
x= 4/2 or x=2/2
Answer: x = 2 and x=1
Giving valid reasons for the steps
Factoring Method
x^2 - 3x + 2 = 0-----(*)
x^2+[(-2x)+(-x)]+2=0 (splitting the mid term as the sum of two quantities so that their product gives the product of the square term and the constant terms of the quadratic expression on the LHS)
(x^2-2x)-x+2=0
x(x-2)-1(x-2)=0 (why pull out (-1)? So as to get (x-2) in the other part too.)
xp-p=0 where p =(x-2)
p(x-1)=0
which is (x-2)(x-1)=0
(x-2) = 0 implies x = 2
(x-1) = 0 implies x = 1
Answer: x = 2 and x=1
Completing the square method
x^2 - 3x + 2 = 0-----(*)
x^2 - 3x = -2
(x)^2-2X(3/2)Xx = -2 [the LHS is of the form a^2-2ab where a = x and b = 3/2
and we need (b^2) to complete the ssquare and hence adding (3/2)^2 = 9/4 to both the sides]
(x)^2-2X(3/2)Xx +(9/4) = -2 +(9/4)
(X-3/2)^2 = (-8+9)/4
(x-3/2)^2 = 1/4
(x-3/2) = (+ or minus)(1/2)
x = +3/2+ or minus(1/2)
That is
x = 3/2+1/2 = 2 or x= 3/2-1/2 = 1
Answer: x = 2 and x = 1
Formula method
x^2 - 3x + 2 = 0-----(*)is of the form (a)x^2 +bx+c = 0
with a = 1, b =(-3) and c= 2
By formula x = {(-b)+or minus[sqrt(b^2-4ac)]}/(2a)}
Therefore x= {-(-3)+ or minus sqrt[(-3)^2-4X1X2]}/(2X1)
x= [3+ or minussqrt(9-8)]/2
x= [3+ or minus 1]/2
x= (3+1)/2 or x= (3-1)/2
x= 4/2 or x=2/2
That is x = 2, x=1
Answer: x = 2 and x=1
Detailed explanation:
Here is the Explanation:(in the form of a conversation!)
consider the quadratic expression on the LHS
x^2-3x+2----(1)
If it is factorisable, then the term in x should conform to getting expressed as the sum of two terms in x so that these two when multiplied should give the product of the square term and the constant term in the expression.
How do you split the term in x in the required way?
Multiply the square term and the constant term
(x^2)*(2) = 2x^2
Find the factors of 2 that you see here in this product
(that is the numerical coefficient of x^2)
2= 1*2
Group the factors of 2 (taking into consideration all the factors and here you have only two factors) and form two parts in such a way that their sum is the coefficient of x.
That is (2) and (1) so that -3 = (-2)+(-1) and (-2)*(-1) = +2
Though you may normally leave out the factor 1 as 1 multiplied by anything is the samething here 1 is very valuable as the total no: of factors is only two!
[if the term in x is negative and the product (of the square term and the constant term) is positive then to both the parts give negative sign so that negative multiplied by negative is positive and negative added to negative is negative]
So, here (-3x) = (-2x)+(-x) and (-2x)*(-x) = 2x^2
x^2-3x+2 becomes
=x^2+(-2x-x)-12
Then group these four terms into a sum of two terms containing a common quantity.This common quantity will be a linear factor in x which we shall call p for convenience for the moment. Take p out. You get p multiplied by another linear factor in x. Put the value of p. Then you get the required answer namely the given expression as a product of two linear factors in x
=(x^2-2x)+(-x+2) by additive associativity
= x(x-2)-1(x-2) (why pull out (-1)? So as to get (x-2) in the other part too.)
= xp-p where p = (x-2)
= p(x-1)
=(x-2)(x-1)
Verification: Putting x=2 in (1)
LHS = x^2 - 3x + 2 = 0-----(*)
=2^2-3X2+2 = 4-6+2 = 0 = RHS
Putting x=1 in (1)
LHS = x^2 - 3x + 2 = 0-----(*)
=1^2-3X1+2 = 1-3+2 = 0 = RHS
Therefore our values are correct
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