# Finally, A New Method for Solving Quadratic Equations That’s Easy to Recall and Apply

May 14 , 2020Quadratic equations have the form, ax^{2}+bx+c=0, were and are given, and are sought. There are three traditional approaches to solving such equations: **applying the quadratic formula**, completing the square, and factoring. Unfortunately, though, students generally find it hard to remember the formula and the steps for completing the square, and they also find the trial-and-error aspect of factoring off-putting. But luckily, a new method has emerged that is easy to recall and involves no hit-or-miss guesswork.

In late 2019, Carnegie Mellon University professor Po-Shen Loh was brainstorming when, to his surprise, he stumbled on a simple and efficient new way to solve quadratic equations. “I was dumbfounded,” he declared. “How can it be that I’ve never seen this before … in any textbook?” And indeed, his new method is so straightforward and intuitive, it’s a minor miracle that it has escaped notice until now.

To illustrate Loh’s method and contrast it against the traditional ones, let’s apply all four to solving the equation x^{2}-2x-48=0.

**1.Solving x ^{2}-2x-48=0 Using the Quadratic Formula**

The quadratic formula represents the solutions to the general quadratic equation ax

^{2}+bx+c=0 as

x= (-b±√(b

^{2}-4ac))/2a. Now you can see why students find this formula intimidating and hard to memorize! Anyway, for this problem, a=1,b=-2, and c= -48, so by the formula, the solutions x are

(-(-2)±√((-2)

^{2}-4(1)(-48)))/(2(1)) = (2±√196)/2 = (2±14)/2 = 1±7 = -6 or 8.

So, we arrived at an answer, but the demerits of the formula were on full display: It taxes memorization and forces complex and cumbersome computations.

**2. Solving x ^{2}-2x-48=0 by Completing the Square**

We will carry out the following four steps to “complete the square” for the given example.

**A) Rewrite the equation as ax**

^{2}+bx=c.x

^{2}-2x=48.

**B) Add the quantity b ^{2}/4a to both sides.**

Here, b

^{2}/4a=(-2)

^{2}/4(1) =1, so we have x

^{2}-2x+1=48+1; ie, x

^{2}-2x+1=49.

**C) Express the left-hand side as a multiple of a perfect square.**

(x-1)^{2}=49.

(Note, Step B completed the left side as a perfect square, which therefore made Step C possible.)

**D) Square root both sides of the re-expressed equation and solve for x.**

√((x-1)^{2} )= √49

x-1= ±7

x=1±7

x= -6 or 8.

Observe that to use this method, students must recall and execute four intricate steps, which is something they find hard to do.

**3. Solving x ^{2}-2x-48=0 by Factoring**

In this approach, an attempt is made to factor the left-hand side of the given quadratic equation by searching for two numbers r and s whose sum is b and product is ac. So, here, we require r+s= -2 and rs=1(-48)= -48.

Now comes the trial-and-error part.

Does r=1 and s= -48 work? No, because while rs= -48,r+s-47 ≠ -2.

How about r=2 and s= -24? Still no, because while rs= -48,r+s-22 ≠ -2.

Several more tries and frustrating moments later, we find that the two numbers that work are r = 6 and s= -8.

Next, we use these two numbers in a process called “successive grouping” to arrive at a factorization of x^{2}-2x-48.

x^{2}-2x-48 = x^{2}+6x-8x-48 (use the fact that b=r+s)

=x(x+6)-8(x+6)=(x-8)(x+6) (from successive grouping).

Last, we use the factored form of the equation to solve.

(x-8)(x+6)=0 ⇒ either x-8=0 or x+6=0 ⇒ either x=8 or x=6.

Closing note: Students generally find the search for r and s extremely exasperating, with good reason!

**4. Solving x ^{2}-2x-48=0 using Professor Loh’s New Method**

Let’s now experience the advantages that Dr. Loh’s method has over the three presented above. Here is his approach:

**A) Compute the quantities**M=(-b)/2a and N=c/a.

Here, M=(-(-2))/2(1) =1 and N=(-48)/1= -48.

Note, this step features two easy-to-remember and easy-to-perform calculations.

**B) Solve the auxiliary equation (M+u)(M-u)=N for u.**

(1+u)(1-u)= -48

1^{2}-u^{2}= -48

49=u^{2}

7=u

Observe, the auxiliary equation in this step is easy to remember and easy to solve.

**C) Quickly recover the solution to the original quadratic equation. They are, simply, the two factors in the auxiliary equation, M+u and M-u.**

x=M+u=1+7=8 or x=M-u=1-7= -6.

Wow, that was almost too easy!

Conclusion

We have compared four methods for solving quadratic equations, three traditional ones, and a new one. We found that the new method, #4, was the fastest, easiest to remember, and easiest to perform. Therefore, students are advised to use it as their go-to technique whenever they must solve a quadratic equation.

**Tags :**
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