# Solving Quadratic Equations

Topics: Quadratic equation, Elementary algebra, Polynomial Pages: 11 (2966 words) Published: April 22, 2013
While the ultimate goal is the same, to determine the value(s) that hold true for the equation, solving quadratic equations requires much more than simply isolating the variable, as is required in solving linear equations. This piece will outline the different types of quadratic equations, strategies for solving each type, as well as other methods of solutions such as Completing the Square and using the Quadratic Formula. Knowledge of factoring perfect square trinomials and simplifying radical expression are needed for this piece. Let’s take a look! Standard Form of a Quadratic Equation

ax2+ bx+c=0

Where a, b, and c are integers
and a≥1

I. To solve an equation in the form ax2+c=k, for some value k. This is the simplest quadratic equation to solve, because the middle term is missing. Strategy: To isolate the square term and then take the square root of both sides. Ex. 1) Isolate the square term, divide both sides by 2

Take the square root of both sides
2x2=40
2x22= 40 2
x2 =20
Remember there are two possible solutions
x2= 20
Simplify radical; Solutions
x= ± 20
x=± 25
(Please refer to previous instructional materials Simplifying Radical Expressions )

II. To solve a quadratic equation arranged in the form ax2+ bx=0. Strategy: To factor the binomial using the greatest common factor (GCF), set the monomial factor and the binomial factor equal to zero, and solve. Ex. 2) 12x2- 18x=0

6x2x-3= 0Factor using the GCF
6x=0 2x-3=0Set the monomial and binomial equal to zero
x=0 x= 32Solutions
* In some cases, the GCF is simply the variable with coefficient of 1. III. To solve an equation in the form ax2+ bx+c=0, where the trinomial is a perfect square. This too is a simple quadratic equation to solve, because it factors into the form m2=0, for some binomial m. (For factoring instructional methods, select The Easy Way to Factor Trinomials ) Strategy: To factor the trinomial, set each binomial equal to zero, and solve. Ex. 3) x2+ 6x+9=0

x+32=0Factor as a perfect square
x+3x+3= 0Not necessary, but valuable step to show two solutions x+3=0 x+3=0Set each binomial equal to zero
x= -3 x= -3Solve
x= -3Double root solution

IV. To solve an equation in the form ax2+ bx+c=0, where the trinomial is not a perfect square, but factorable. Similar to the last example, this is a simple quadratic equation to solve, because it factors into the form mn=0, for some binomials m and n. Strategy: To factor the trinomial, set each binomial equal to zero, and solve. Ex. 4) 2x2-x-6=0

* Using the factoring method from The Easy Way to Factor Trinomials, we need to find two number that multiply to give ac, or -12, and add to give b, or -1. These values are -4 and 3. Rewrite the trinomial with these two values as coefficients to x that add to the current middle term of -1x.

2x2- 4x+3x-6=0Rewrite middle term
2x2- 4x+3x-6=0

2xx-2+ 3x-2= 0Factor by grouping
x-22x+3= 0Factor out the common binomial (x-2)
x-2=0 2x+3=0Set each binomial equal to zero
x=2 x= -32Solutions

V. To solve a quadratic equation not arranged in the form ax2+ bx+c=0, but factorable.
Strategy: To combine like terms to one side, set equal to zero, factor the trinomial, set each
binomial equal to zero, and solve.
Ex. 5) 6x2+ 2x-3=9x+2
-9x -9x
6x2- 7x-3= 2
-2 -2
6x2- 7x-5=0
* To factor this trinomial, we are looking for two numbers that multiply to give ac, or -30, and add to give b, or -7. These values would be 3 and -10. Rewrite the trinomial with these two values as coefficients to x that add to the current middle...

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