In standard form, quadratic equations take the form y=ax2+bx+c, and can be solved in various ways. In general, they are solved for the value(s) of x that make the equation equal to 0. Thus, y=ax2+bx+c becomes 0=ax2+bx+c. Graphically, all points with a y-coordinate of 0 are the x-intercepts of the function — or the zeros of the parabola. That means, solving a quadratic equation leads to finding the zeros of the parabola. Since a parabola can have 0,1, or 2 zeros, a quadratic equation can have 0,1, or 2 solutions.
If neither side of the equation equals 0, rearrange it so that all terms are on the same side. 2x2+3x−5=5x+7⇔2x2+3x−5−5x−7=0 If there is more than one term of the same degree, simplify the equation by combining like terms. 2x2+3x−5−5x−7=0⇔2x2−2x−12=0
The side with the variables can now be seen as a function, f(x). Plot that function in a coordinate plane. For the equation 2x2−2x−12=0, the function is f(x)=2x2−2x−12.
Now, find all the points on the graph that have the y-coordinate 0.
The x-coordinates of any identified points solve the equation, f(x)=0.
In the example, the x-coordinates are -2 and 3. Note that solving an equation graphically does not necessarily lead to an exact answer. To verify a solution, substitute it into f(x) and evaluate the expression. Verifying the solution x=-2 is done by evaluating f(-2).
Simple quadratic equations take the form ax2+c=0 and are solved using inverse operations. Once x2 remains on the left-hand side, the equation can be written as x2=d, where d=a-c. The value of d gives the number of solutions the equation has.
d>0d=0d<0:2 real solutions:1 real solution:0 real solutionsseeits factors. 4x2+13x+3 This expression can be factored by finding a pair of integers whose product is a⋅c, which here is 4⋅3, and whose sum is b, which in the example is 13.
The first step is to find all possible pairs of integers that multiply to a⋅c. In this case, a=4 and c=3. Thus, their product is 3⋅4=12. To find all factor pairs, start with the pair where one factor is 1. The other factor must then be 12. Then continue with the pair where one of the factors is 2, and so forth. In this case, there are three pairs. 1and122and63and4
If the given expression is factorable, one of the factor pairs will add to equal b. In this case, b=13. 1+122+63+4=13=8=7 Here, 1 and 12 is the only factor pair that adds to 13.
Now, use the factor pair to rewrite the x-term of the original expression as a sum. Since the factor pair is 1 and 12, the middle term can be written as 13x=12x+x. This gives the following equivalent expression. 4x24x2+ 13x + 12x+x+3+3 Notice that the expression hasn't been changed. Rather, it has been rewritten.
Now, the expression has four terms, which can be grouped into the first two terms and the last two terms. Then, the GCF of each group can be factored out. Here, begin with the first group of terms, 4x2 and 12x. Notice how each can be written as a product of its factors. 4x2 4⋅x⋅x + 12x+ 4⋅3⋅x The GCF is 4x. Therefore, it's possible to factor out 4x. 4x2+12x+x+34x(x+3)+x+3
Next, repeat the same process with the last two terms. In this case, x and 3 do not have any common factors, but it's always possible to write expressions as a product of 1 and itself. 4x(x+3)4x(x+3)+x+3+1(x+3)
If all previous steps have been performed correctly, there should now be two terms with a common factor, which can be seen as a repeated parentheses. Factoring (x+3) out of both terms gives 4x(x+3)+(x+3)(41(x+3)x+1). This means that 4x2+13x+3 can be written in factored form as (x+3)(4x+1).
Solve the following equation by factoring. x2−2x−3=0