{"id":9532,"date":"2026-06-01T21:33:48","date_gmt":"2026-06-01T21:33:48","guid":{"rendered":"https:\/\/kapdec.com\/help\/?p=9532"},"modified":"2026-06-02T22:54:54","modified_gmt":"2026-06-02T22:54:54","slug":"techniques-of-solving-quadratic-equations","status":"publish","type":"post","link":"https:\/\/kapdec.com\/help\/techniques-of-solving-quadratic-equations\/","title":{"rendered":"Techniques Of Solving Quadratic Equations"},"content":{"rendered":"\n\n\n
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Unit: Quadratic Equations<\/strong><\/h2>\n

Techniques of Solving Quadratic Equations<\/strong><\/h2>\n

Splitting middle term:<\/strong><\/p>\n

There is a method that works in simple cases. With the quadratic equation in this form:<\/p>\n

ax<\/strong>2<\/sup><\/strong> + bx + c = 0<\/strong><\/p>\n

Step1: <\/strong> Find two numbers that multiply to give ac (in other words a times c), and add to give b.<\/p>\n

Example: x2<\/sup> + 4x – 12 = 0<\/p>\n

ac is -12 x 1= 6 and b is 4<\/p>\n

So, we want two numbers that multiply together to make -12, and there difference is 4.<\/p>\n

In fact, 6 and -2 do that (6 x – 2 = -12 and 6 – 2 = 4)<\/p>\n

Step2: <\/strong>Rewrite the middle term with those numbers:<\/p>\n

Rewrite 4x with 6x and -2x<\/p>\n

x2<\/sup> + 6x – 2x – 12<\/p>\n

Step3: <\/strong>Factor the first two and last two terms separately.<\/p>\n

The first two terms x2<\/sup> + 6x factor into x(x + 6)<\/p>\n

The last two terms -2x – 12 factors into -2(x + 6)<\/p>\n

So, we get:<\/p>\n

x(x+6) – 2(x + 6)<\/p>\n

Step4: <\/strong>If we have done this correctly, our two new terms should have a clearly visible common factor.<\/p>\n

In this case we can see that (x + 6) is common to both terms, so we can write above calculation as follows:<\/p>\n

Start with: x(x+6) – 2(x + 6)<\/p>\n

(x – 2)( x + 6)<\/p>\n

(x – 2) and (x – 6) are factors of quadratic equation x2<\/sup> + 4x – 12 = 0.<\/p>\n

Here,<\/p>\n

(x – 2)(x + 6) = 0<\/p>\n

So, (x – 2) = 0<\/p>\n

x = 2<\/p>\n

And (x + 6) = 0<\/p>\n

x = -6<\/p>\n

Therefore, 2 and -6 are root or zero of the given equation.<\/p>\n

In general, a real number ∝<\/em>\"\" is called a root of the quadratic equation<\/p>\n

ax2<\/sup> + bx + c = 0, a ≠<\/em> 0 if a \"\"2 + b∝<\/em> + c = 0. We also say that x = <\/strong>∝<\/em>is a solution of the quadratic equation<\/strong>, or that ∝<\/em>satisfies the quadratic equation<\/strong>. Note that the zeroes of the quadratic polynomial ax2 <\/sup>+ bx + c and the roots of the quadratic equation ax2<\/sup> + bx + c = 0 are the same<\/strong>.<\/p>\n

 <\/p>\n

How to Complete the Square<\/strong><\/p>\n

Let’s start with a general polynomial equation of degree 2:<\/p>\n

x2<\/sup> – x + 10 = -x2<\/sup> – 17x + 28<\/a><\/p>\n

This time, instead of moving all terms to one side of the equation, we will have all terms with variables on one side of the equation and all constants on the other side:<\/p>\n

x2<\/sup> – x + 10 + x2<\/sup> + 17x – 10 = -x2<\/sup> – 17x + 28 + x2<\/sup> + 17x – 10<\/p>\n

2x2<\/sup> + 16x = 18<\/p>\n

In any completing the square problem, there can be no coefficient before the x2<\/sup>. If there is one, all terms must be divided by that coefficient. In this case, we will divide all terms by 2:<\/p>\n

2x2<\/sup> ÷ 2 + 16x ÷ 2 = 18 ÷ 2<\/p>\n

x2<\/sup> + 8x = 9<\/p>\n

For any completing the square problem, we must reduce the equation to the form x2<\/sup> + bx = c. For the next step, we should keep in mind the following identity:<\/p>\n

(x + y)2<\/sup> = x2<\/sup> + 2xy + y2<\/sup><\/p>\n

Now, let’s assume b = 2y. That would mean x2<\/sup> + bx is almost (x + \"\")2<\/sup>. All it would be missing is \"\". Therefore, the next step is to add \"\" to both sides of the equation. In the case of our example, we will add     \"\" = 16 to both sides:<\/p>\n

x2<\/sup> + 8x + 16 = 9 + 16<\/p>\n

x2<\/sup> + 8x + 16 = 25<\/p>\n

We know from the identity (x + y)2<\/sup> = x2<\/sup> + 2xy + y2<\/sup> that we can now condense x2<\/sup> + 8x + 16 to (x + 4)2<\/sup>:<\/p>\n

(x + 4)2<\/sup> = 25<\/p>\n

Now you can take the square root of both sides; keep in mind this is both the positive and the negative square root in this case:<\/p>\n

\"\"<\/p>\n

The solutions to the equation x2<\/sup> – x + 10 = -x2<\/sup> – 17x + 28 are x = 1 and x = -9. To check our answer, let’s see what happens if we expand the factors (x – 1) and (x + 9):<\/p>\n

(x – 1)(x + 9) = 0<\/p>\n

x(x + 9) – x – 9 = 0<\/p>\n

x2<\/sup> + 8x = 9<\/p>\n

This leads us right back to the equation we had earlier, showing that we get the same answers through factoring and through completing the square.<\/p>\n

In summary, go through the following steps to complete the square:<\/p>\n