Quadratic Equation Methods Quadratic Equation Easy Method

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A quadratic equation is a polynomial equation in a single variable where the highest exponent of the variable is 2.^[1] There are three main ways to solve quadratic equations: 1) to factor the quadratic equation if you can do so, 2) to use the quadratic formula, or 3) to complete the square. If you want to know how to master these three methods, just follow these steps.

1. 1

2. 2

   Factor the expression. To factor the expression, you have to use the factors of the ${\displaystyle x^{2}}$ term (3), and the factors of the constant term (-4), to make them multiply and then add up to the middle term, (-11). Here's how you do it:

   • Since ${\displaystyle 3x^{2}}$ only has one set of possible factors, ${\displaystyle 3x}$ and ${\displaystyle x}$ , you can write those in the parenthesis: ${\displaystyle (3x\pm ?)(x\pm ?)=0}$ .
   • Then, use the process of elimination to plug in the factors of 4 to find a combination that produces -11x when multiplied. You can either use a combination of 4 and 1, or 2 and 2, since both of those numbers multiply to get 4. Just remember that one of the terms should be negative, since the term is -4.^[3]
   • By trial and error, try out this combination of factors ${\displaystyle (3x+1)(x-4)}$ . When you multiply them out, you get ${\displaystyle 3x^{2}-12x+x-4}$ . If you combine the terms ${\displaystyle -12x}$ and ${\displaystyle x}$ , you get ${\displaystyle -11x}$ , which is the middle term you were aiming for. You have just factored the quadratic equation.
   • As an example of trial and error, let's try checking a factoring combination for ${\displaystyle 3x^{2}-11x-4=0}$ that is an error (does not work): ${\displaystyle (3x-2)(x+2)}$ = ${\displaystyle 3x^{2}+6x-2x-4}$ . If you combine those terms, you get ${\displaystyle 3x^{2}-4x-4}$ . Though the factors -2 and 2 do multiply to make -4, the middle term does not work, because you needed to get ${\displaystyle -11x}$ , not ${\displaystyle -4x}$ .

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3. 3

   Set each set of parenthesis equal to zero as separate equations. This will lead you to find two values for ${\displaystyle x}$ that will make the entire equation equal to zero, ${\displaystyle (3x+1)(x-4)}$ = 0. Now that you've factored the equation, all you have to do is put the expression in each set of parenthesis equal to zero. But why? -- because to get zero by multiplying, we have the "principle, rule or property" that one factor must be zero, then at least one of the factors in parentheses, as ${\displaystyle (3x+1)(x-4)}$ must be zero; so, either (3x + 1) or else (x - 4) must equal zero. So, you would write ${\displaystyle 3x+1=0}$ and also ${\displaystyle x-4=0}$ .

4. 4

   Solve each "zeroed" equation independently. In a quadratic equation, there will be two possible values for x. Find x for each possible value of x one by one by isolating the variable and writing down the two solutions for x as the final solution. Here's how you do it:

   • Solve 3x + 1 = 0
     • 3x = -1 ..... by subtracting
     • 3x/3 = -1/3 ..... by dividing
     • x = -1/3 ..... simplified
   • Solve x - 4 = 0
     • x = 4 ..... by subtracting
   • x = (-1/3, 4) ..... by making a set of possible, separate solutions, meaning x = -1/3, or x = 4 seem good.

5. 5

   Check x = -1/3 in (3x + 1)(x – 4) = 0:

   We have

   (3[-1/3] + 1)([-1/3] – 4) ?=? 0

   ..... by substituting (-1 + 1)(-4 1/3) ?=? 0 ..... by simplifying (0)(-4 1/3) = 0 ..... by multiplying therefore 0 = 0 ..... Yes, x = -1/3 works

6. 6

   Check x = 4 in (3x + 1)(x - 4) = 0:

   We have (3[4] + 1)([4] – 4) ?=? 0

   ..... by substituting (13)(4 – 4) ?=? 0 ..... by simplifying (13)(0) = 0 ..... by multiplying 0 = 0 ..... Yes, x = 4 works

   • So, both solutions do "check" separately, and both are verified as working and correct for two different solutions.

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1. 1

2. 2

   Write down the quadratic formula. The quadratic formula is: ${\displaystyle {\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}}$ ^[5]

3. 3

   Identify the values of a, b, and c in the quadratic equation. The variable a is the coefficient of the x² term, b is the coefficient of the x term, and c is the constant. For the equation 3x² -5x - 8 = 0, a = 3, b = -5, and c = -8. Write this down.

4. 4

   Substitute the values of a, b, and c into the equation. Now that you know the values of the three variables, you can just plug them into the equation like this:

   • {-b +/-√ (b² - 4ac)}/2
   • {-(-5) +/-√ ((-5)² - 4(3)(-8))}/2(3) =
   • {-(-5) +/-√ ((-5)² - (-96))}/2(3)

5. 5

   Do the math. After you've plugged in the numbers, do the remaining math to simplify positive or negative signs, multiply, or square the remaining terms. Here's how you do it:

   • {-(-5) +/-√ ((-5)² - (-96))}/2(3) =
   • {5 +/-√(25 + 96)}/6
   • {5 +/-√(121)}/6

6. 6

   Simplify the square root. If the number under the radical symbol is a perfect square, you will get a whole number. If the number is not a perfect square, then simplify to its simplest radical version. If the number is negative, and you're sure it's supposed to be negative, then the roots will be complex. In this example, √(121) = 11. You can write that x = (5 +/- 11)/6.

7. 7

   Solve for the positive and negative answers. If you've eliminated the square root symbol, then you can keep going until you've found the positive and negative results for x. Now that you have (5 +/- 11)/6, you can write two options:

   • (5 + 11)/6
   • (5 - 11)/6

8. 8

   Solve for the positive and negative answers. Just do the math:

   • (5 + 11)/6 = 16/6
   • (5-11)/6 = -6/6

9. 9

   Simplify. To simplify each answer, just divide them by the largest number that is evenly divisible into both numbers. Divide the first fraction by 2, and divide the second by 6, and you have solved for x.

   • 16/6 = 8/3
   • -6/6 = -1
   • x = (-1, 8/3)

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1. 1

   Move all of the terms to one side of the equation. Make sure that the a or x² term is positive. Here's how you do it:^[6]

   • 2x² - 9 = 12x =
   • 2x² - 12x - 9 = 0
     • In this equation, the a term is 2, the b term is -12, and the c term is -9.

2. 2

   Move the c term or constant to the other side. The constant term is the numerical term without a variable. Move it to the right side of the equation:

   • 2x² - 12x - 9 = 0
   • 2x² - 12x = 9

3. 3

   Divide both sides by the coefficient of the a or x² term. If x² has no term in front of it, and just has a coefficient of 1, then you can skip this step. In this case, you'll have to divide all of the terms by 2, like so:

   • 2x²/2 - 12x/2 = 9/2 =
   • x² - 6x = 9/2

4. 4

   Divide b by two, square it, and add the result to both sides. The b term in this example is -6. Here's how you do it:

   • -6/2 = -3 =
   • (-3)² = 9 =
   • x² - 6x + 9 = 9/2 + 9

5. 5

   Simplify both sides. Factor the terms on the left side to get (x-3)(x-3), or (x-3)². Add the terms on the right side to get 9/2 + 9, or 9/2 + 18/2, which adds up to 27/2.

6. 6

   Find the square root of both sides. The square root of (x-3)² is simply (x-3). You can write the square root of 27/2 as ±√(27/2). Therefore, x - 3 = ±√(27/2).

7. 7

   Simplify the radical and solve for x. To simplify ±√(27/2), look for a perfect square within the numbers 27 or 2 or in their factors. The perfect square 9 can be found in 27, because 9 x 3 = 27. To take 9 out of the radical sign, pull out the number 9 from the radical, and write the number 3, its square root, outside the radical sign. Leave 3 in the numerator of the fraction under the radical sign, since that factor of 27 cannot be taken out, and leave 2 on the bottom. Then, move the constant 3 on the left side of the equation to the right, and write down your two solutions for x:

   • x = 3 + 3(√6)/2
   • x = 3 - 3(√6)/2)

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Add New Question

• Question

  What is the fastest way to solve a simple quadratic equation?

  

  David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor's degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math.

  

  Academic Tutor

  Expert Answer

  Use the square root method! If the only variable in your equation is x², move all of the other numbers to the other side of the equation. Then, find the square root of both sides of the equation.

• Question

  I am totally lost. Are than any really simple shortcuts for one to use?

  

  Sorry, no. Factoring quadratics is definitely a challenge for many people. Console yourself with the knowledge that you're not likely ever to need this skill in real life. Passing your next math test is another matter.

• Question

  How do I solve the following equation: Find the values of k for which the roots of the equation (k+4)x^2+(k+1)x+1=0 are real and equal?

  

  Having a double root means that your espression is a perfect square. It also means that the discriminate is zero. Using either of these will give an equation in k that you can solve for the answer. A square takes the form (ax+b)^2= a^2 x^2 + 2abx + b^2. Obviously, then b = +1 or -1, leaving a=+/- (k+1)/2 = sqrt(k+4). Setting the discriminant of the original quadratic to zero gives (k+1)^2 - 4 (k+4) = 0. Whichever way you prefer you should get k=-3 (x^2 -2x +1 = (x-1)^2) and k=5 (9x^2 + 6x +1 = (3x+1)^2).

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• If the number under the square root is not a perfect square, then the last few steps run a little differently. Here is an example:^[7]

• As you can see, the radical sign did not disappear completely. Therefore, the terms in the numerator cannot be combined (because they are not like terms). There is no purpose, then, to splitting up the plus-or-minus. Instead, we divide out any common factors --- but ONLY if the factor is common to both of the constants AND the radical's coefficient.

• If the "b" is an even number, the formula is : {-(b/2) +/- √(b/2)-ac}/a.

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Article Summary X

To solve quadratic equations, start by combining all of the like terms and moving them to one side of the equation. Then, factor the expression, and set each set of parentheses equal to 0 as separate equations. Finally, solve each equation separately to find the 2 possible values for x. To learn how to solve quadratic equations using the quadratic formula, scroll down!

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