3.3 Solving Quadratics by Factoring
First, if we want to begin solving quadratics by factoring, we must know how to factor as it is the basics of solving quadratic equations. For example, if you want to solve (2x² – 15x + 25 = 0), we must begin with factoring this equation, and that is by identifying the product of A and C and finding what two numbers make that product of A and C that also is the sum of B. Such as, in this case, A = 2, B = -15, and C =25, there are two methods in factoring this which is either the X method or the square box method, personally I prefer the X method. The factors of 50 (because A • C = 2 • 25 = 50) is 1 • 50, 2 • 25, and 5 • 10. Which one of these has a difference or sum of -15? 5 and 10, in this case, it would be -5 and -10 because (-5 • -10 = our (ac) value of 50 as well as it is the sum of our (b) value of -15 (-5 – -10 = -15).) Next, once we have these values we can use the X method to factor,
after we have placed our values, we can divide the left and right side by the A value
Yes the 1 underneath the 5 is unnecessary but personally, it helps me visualize the factor more easily. Then, to factor it is (2x-5)(x-5), personally, how I usually distinguish which numbers go where is if you look at the numerator of each fraction which is (-5 and -5) you can recognize the plus or minus symbol which is a minus in this case and if you compare it to the recipe of a factored quadratic which looks like (x -/+ #)(x -/+ #), you can sort of visualize the placement of the numerator value into a factored form from matching the symbol (+ -) placement. Another way to remember is bottoms up which is to find the denominator (bottom) and moving it up which makes it first and before the numerator when writing them into two factors in brackets. After that, we can find which values would make it so the equation would equal 0. For (2x-5)(x-5) X = 5/2 and X = 5 because (2 (5/2) – 5 = 0) and (5-5 = 0)
Once we know how to factor we can finally begin solving quadratics. A question that looks different but uses the same formula of factoring quadratics is (2x – 3)(x + 1) = 3. First, the right side must equal 0 and the left side must be in general form (Ax²+Bx+C), so expand the left side and simplify,
2x² + 2x – 3x -3 -3 = 0
2x² – x – 6 = 0
Factor trinomial
(2x + 3)(x – 2) = 0
Then, to solve, 2x + 3 = 0 -> 2x = -3 -> x = -3/2
AND x – 2 = 0 -> x = 2
After we can verify the solution,
(2(2) -3)(2+ 1) = 3
(1)(3) = 3
3 = 3
AND
(2(-3/2) – 3) (-3/2 +1) = 3
(-6)(-1/2) = 3
3 = 3
Next, we can try an equation that involves a common factor, one example could be (2x² + 18 = 12x)
First we must move everything to the left side so the equation could equal 0
(2x² -12x +18 = 0)
Then remove GCF which is 2 because the value of A, B, C in the equation can be divided by 2 (2,-12,18)
2(x² -6x + 9) = 0
Then factor
2(x – 3) (x – 3) = 0
X = 3, X = 3
Verify
2(3)² + 18 = 12(3)
18 + 18 = 36
36 = 36
One more simple example that is not quite the general form but still uses the same properties of solving quadratic equations is
(2x²=4x)
Move 4x to the left side to make the equation equal to 0
(2x² – 4x = 0)
Factor the polynomial
(2x (x – 2) = 0)
X = 2 and X = 0 (x = 0 because if you put 0 in the x in 2x then it would be 2(0) and 2 • 0 = 0)
Verify
2(2)² = 4(2)
8 = 8
AND
2(0)² = 4(0)
0 = 0
You can also apply this strategy of solving quadratic with radical equations, such as (√23-x = x – 3), we must rewrite this equation into Ax²+Bx+C = 0 form, to do that we first isolate the radical, which makes the equation to,
(√23-x)² = (x – 3)²
we square both sides to remove radical
(23 – x = (x – 3)(x – 3) )
Then expand the right side
(23 – x = x² -6x + 9)
Move the left side to the right side to make the equation equal 0
(x² – 5x – 14 = 0)
Factor
(x – 7) (x + 2)
X = 7, and x = -2
Verify
(√23- (7) = (7) – 3)
(√16 = 4)
4 = 4
AND
(√23- (-2) = (-2) – 3)
(√25 = -5)
5 ≠ -5
As you can see all the questions before this the solution worked when verifying, however, when it comes to radicals the solution may not always work, which is why it is very important to verify especially when dealing with radicals. This means that (-2) in the example above is an extraneous root, where it may seem like the solution would work because you factored it, but it actually doesn’t and is false.
An example that uses the method of solving quadratic but also incorporates a method of the previous chapter of substituting variables is (2x-1)² – 2(2x-1) – 8 = 0, this looks like Ax² + Bx + C = 0, we can replace (2x-1) to W, which makes the equation w² – 2w – 8 = 0 then we factor, (w-4)(w+2), then we replace (w) with (2x-1), which equals to (2x-1-4)(2x-1+2), simplify, (2x-5)(2x+1), then x = 5/2 and x = -1/2, verify, also when verifying, make sure you input the solutions in the original equation which is (2x-1)² – 2(2x-1) – 8 = 0, to make sure the answers are accurate
X=5/2:(4)² – 2(4) – 8 = 0, 0 = 0, AND X=-1/2: (-2)² – 2(-2) – 8 = 0, 0 = 0.
Lastly, we can use a quadratic equation to solve a problem, this makes you think differently as you need to know which numbers are needed and where they are needed to be put, instead of just doing equations already given to you.
Such as, the area of a rectangular sheep pen is 96m². The pen is divided into two smaller pens by inserting a fence parallel to the width of the pen. A total of 48 m of fencing is used. Determine the dimensions of the pen.
The area of the pen is 96m² and the width is W in metres, to find the length it would be 96/W = L, which are the dimensions. The total length of the fence is 48m so an equation would be (3w + 2(96/w) = 48) First we multiply each term by w to remove the w underneath the 96, (3w² + 2(96) = 48w), make it general form, (3w² -48w + 192), next we divide each term by 3, 3(w² – 16 + 64), then factor, 3(w – 8)(w – 8), so (w = 8), lastly 96m²(area)/ 8m(width) = 12m(length), so the sheep pen has a length of 12m, a width of 8m, and a area of 96m²
In conclusion, solving quadratic by factoring is simple and some important things to remember and to do is when solving equations are, know how to factor quadratics, know how to make it into general form (Ax² + Bx + C= 0), verify the solution (especially radicals), and your main goal is to know X or what the variable equals to; to make the equation true.