{"id":36431,"date":"2024-07-10T15:58:20","date_gmt":"2024-07-10T07:58:20","guid":{"rendered":"https:\/\/www.wukongsch.com\/blog\/?p=36431"},"modified":"2026-01-12T17:42:16","modified_gmt":"2026-01-12T09:42:16","slug":"quadratic-equation","status":"publish","type":"post","link":"https:\/\/www.wukongsch.com\/blog\/quadratic-equation-post-36431\/","title":{"rendered":"Quadratic Equation: Formulas, Solutions, Graphs, and Examples"},"content":{"rendered":"
<\/div>\n

Are you looking to demystify the world of quadratic equations<\/strong>? Whether you’re a student, educator, or simply curious about mathematics, this guide offers a thorough exploration of quadratic equations, including their definition, quadratic formula<\/strong>, graphical representations, and practical applications. Dive in to discover how these equations are integral to various fields and how you can solve them with ease.<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

<\/span>Part 1. What is the Quadratic Equation?<\/span><\/h2>\n\n\n\n

Definition of Quadratic Equation<\/h3>\n\n\n\n

The quadratic equation is central to algebra, and it arises in many forms throughout mathematics and its applications. A quadratic equation is defined by the largest power of the variable, which is squared. The conventional form is ax^2 +bx +c =0<\/p>\n\n\n\n

where ‘a’ does not equal 0. The roots of this equation represent the values of ‘x’ that satisfy the equation. It’s noteworthy to note that a quadratic equation always has two roots, which can be real or complex, depending on the discriminant.<\/p>\n\n\n

\r\n
\r\n
<\/div>\r\n

Discovering the maths whiz in every child,
\nthat’s what we do.<\/span><\/p>\n<\/h3>\r\n

Suitable for students worldwide, from grades 1 to 12.<\/p>\r\n \r\n Get started free!\r\n <\/a>\r\n <\/div>\r\n

<\/div>\r\n<\/div>\n\n\n

In our daily lives, quadratic equations are more present than we might think. From calculating the area of rooms to determining the profit of a product in business, or even finding the speed of an object in motion, quadratic equations play a crucial role.<\/p>\n\n\n\n

Standard Form of Quadratic Equation<\/h3>\n\n\n\n

The standard form of a quadratic equation is :<\/p>\n\n\n\n

\"Standard<\/figure>\n\n\n\n

where ‘a’, ‘b’, and ‘c’ are coefficients and ‘a’ does not equal zero. This form is crucial for solving quadratic equations and understanding their properties.<\/p>\n\n\n\n

<\/span>Part 2. Quadratic Formula<\/span><\/h2>\n\n\n\n

The quadratic formula is a fundamental method for solving quadratic equations. It gives a basic approach for finding the roots of any quadratic equation. To obtain the quadratic formula, we can complete the square for the general form of the quadratic equation. This process leads us to the well-known formula:<\/p>\n\n\n\n

x = [-b \u00b1 \u221a(b((2)) – 4ac)]\/2a.<\/p>\n\n\n

\n
\"Quadratic<\/figure><\/div>\n\n\n

Anyone who wants to solve quadratic equations must know how to apply this formula. It is also critical to understand additional quadratic equation ormulas , such as the sum and product of root formulas or the vertex form.<\/p>\n\n\n

\r\n \r\n Learn more about Quadratic Formula\r\n <\/a>\r\n<\/div>\n\n\n

How to Derivate Quadratic Formula<\/h3>\n\n\n\n

To derive the quadratic formula, we start with the standard form of a quadratic equation: ax^2 + bx + c = 0. The goal is to solve x by isolating it on one side of the equation. Here’s the step-by-step process:<\/p>\n\n\n\n

How to Derivate Quadratic Formula<\/strong><\/td>Steps<\/strong><\/td>Note<\/strong><\/td><\/tr>
Step 1: Subtract c from both sides<\/strong><\/td>ax^2 + bx + c = 0ax^2 + bx = -c<\/td>We want to set up the left side for completing the square, so we move the constant term ‘c’ to the right side.<\/td><\/tr>
Step 2: Divide both sides by a<\/strong><\/td>ax^2 + bx = -cx^2 + (b\/a)x = -c\/a<\/td>Dividing by ‘a’ simplifies the equation and sets us up for the next step.<\/td><\/tr>
Step 3: Complete the square<\/strong><\/td>To complete the square, we take half of the coefficient of x (which is b\/a), square it, and add it to both sides.x^2 + (b\/a)x + (b\/2a)^2 = -c\/a + (b\/2a)^2<\/td>Adding (b\/2a)^2 to both sides allows us to create a perfect square trinomial on the left side.<\/td><\/tr>
Step 4: Simplify the right side<\/strong><\/td>x^2 + (b\/a)x + (b^2\/4a^2) = (-c + b^2\/4a)\/a<\/td>We combine like terms on the right side to simplify the equation.<\/td><\/tr>
Step 5: Write the left side as a perfect square<\/strong><\/td>The left side is now a perfect square trinomial, which can be written as (x + b\/2a)^2. The right side simplifies to (b^2 – 4ac)\/(4a^2).(x + b\/2a)^2 = (b^2 – 4ac)\/(4a^2)<\/td>The left side is now a square of a binomial, which makes it easier to solve for x.<\/td><\/tr>
Step 6: Take the square root of both sides<\/strong><\/td>x + b\/2a = \u00b1\u221a(b^2 – 4ac)\/(2a)<\/td>Taking the square root of both sides allows us to solve for x. The \u00b1 sign indicates that there are two possible solutions.<\/td><\/tr>
Step 7: Solve for x<\/strong><\/td>Finally, we subtract b\/2a from both sides to isolate x:x = [-b \u00b1 \u221a(b^2 – 4ac)]\/(2a)<\/td>This is the quadratic formula, which gives us the roots of the quadratic equation.<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Example:<\/strong><\/p>\n\n\n\n

Let’s derive the quadratic formula for the equation x^2 + 5x + 6 = 0.<\/p>\n\n\n\n

    \n
  1. x^2 + 5x + 6 = 0<\/li>\n\n\n\n
  2. x^2 + 5x = -6<\/li>\n\n\n\n
  3. x^2 + 5x + (5\/2)^2 = -6 + (5\/2)^2<\/li>\n\n\n\n
  4. x^2 + 5x + 25\/4 = -6 + 25\/4<\/li>\n\n\n\n
  5. (x + 5\/2)^2 = 1\/4<\/li>\n\n\n\n
  6. x + 5\/2 = \u00b1\u221a(1\/4)<\/li>\n\n\n\n
  7. x = -5\/2 \u00b1 1\/2<\/li>\n<\/ol>\n\n\n\n

    So the solutions are x = -2 and x = -3.<\/p>\n\n\n\n

    This process demonstrates how the quadratic formula is derived and how it can be applied to find the roots of a quadratic equation.<\/p>\n\n\n\n

    Formulas Related to Quadratic Equations<\/h3>\n\n\n\n
      \n
    1. \n

      Start with a quadratic equation in the form of a<\/mi>x<\/mi>2<\/mn><\/msup>+<\/mo>b<\/mi>x<\/mi>+<\/mo>c<\/mi>=<\/mo>0<\/mn><\/mrow>ax^2 + bx + c = 0<\/annotation><\/semantics><\/math><\/span><\/span>a<\/span>x<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>+<\/span><\/span><\/span><\/span>b<\/span>x<\/span><\/span>+<\/span><\/span><\/span><\/span>c<\/span><\/span>=<\/span><\/span><\/span><\/span>0<\/span><\/span><\/span><\/span><\/span><\/strong>. Ensure that a<\/mi><\/mrow>a<\/annotation><\/semantics><\/math><\/span><\/span>a<\/span><\/span><\/span><\/span><\/span>, the coefficient of x<\/mi>2<\/mn><\/msup><\/mrow>x^2<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>, is 1. If it’s not, divide every term by a<\/mi><\/mrow>a<\/annotation><\/semantics><\/math><\/span><\/span>a<\/span><\/span><\/span><\/span><\/span> to make it 1.<\/p>\n<\/li>\n
    2. \n

      Move the constant term to the other side of the equation<\/strong>. Do this by adding or subtracting the constant term from both sides of the equation so that you have only the x<\/mi>2<\/mn><\/msup><\/mrow>x^2<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> and x<\/mi><\/mrow>x<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span><\/span><\/span><\/span> terms on one side and the constant on the other.<\/p>\n<\/li>\n
    3. \n

      Take half of the coefficient of x<\/mi><\/mrow>x<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span><\/span><\/span><\/span> (which is b<\/mi><\/mrow>b<\/annotation><\/semantics><\/math><\/span><\/span>b<\/span><\/span><\/span><\/span><\/span>) and square it<\/strong>. The coefficient of x<\/mi><\/mrow>x<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span><\/span><\/span><\/span> at this point should be b<\/mi><\/mrow>b<\/annotation><\/semantics><\/math><\/span><\/span>b<\/span><\/span><\/span><\/span><\/span> because we’ve ensured a<\/mi>=<\/mo>1<\/mn><\/mrow>a = 1<\/annotation><\/semantics><\/math><\/span><\/span>a<\/span><\/span>=<\/span><\/span><\/span><\/span>1<\/span><\/span><\/span><\/span><\/span>. So, you take b<\/mi>2<\/mn><\/mfrac><\/mrow>frac{b}{2}<\/annotation><\/semantics><\/math><\/span><\/span><\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>b<\/span><\/span><\/span><\/span><\/span>&ZeroWidthSpace;<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> and square it, giving you (<\/mo>b<\/mi>2<\/mn><\/mfrac>)<\/mo><\/mrow>2<\/mn><\/msup><\/mrow>left(frac{b}{2}right)^2<\/annotation><\/semantics><\/math><\/span><\/span>(<\/span><\/span><\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>b<\/span><\/span><\/span><\/span><\/span>&ZeroWidthSpace;<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>)<\/span><\/span><\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>.<\/p>\n<\/li>\n
    4. \n

      Add this value to both sides of the equation<\/strong>. This step is crucial because it “completes the square” on the left side of the equation.<\/p>\n<\/li>\n

    5. \n

      Rewrite the left side of the equation as a perfect square<\/strong>. You do this by writing it in the form (<\/mo>x<\/mi>+<\/mo>b<\/mi>2<\/mn><\/mfrac>)<\/mo><\/mrow>2<\/mn><\/msup><\/mrow>left(x + frac{b}{2}right)^2<\/annotation><\/semantics><\/math><\/span><\/span>(<\/span><\/span>x<\/span><\/span>+<\/span><\/span><\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>b<\/span><\/span><\/span><\/span><\/span>&ZeroWidthSpace;<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>)<\/span><\/span><\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>.<\/p>\n<\/li>\n
    6. \n

      Solve for x<\/mi><\/mrow>x<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span><\/span><\/span><\/span><\/strong>. This involves taking the square root of both sides, remembering to consider both the positive and negative roots, and then simplifying to find the values of x<\/mi><\/mrow>x<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span><\/span><\/span><\/span>.<\/p>\n<\/li>\n<\/ol>\n

      Let’s go through an example:<\/p>\n

      Solve x<\/mi>2<\/mn><\/msup>+<\/mo>6<\/mn>x<\/mi>+<\/mo>5<\/mn>=<\/mo>0<\/mn><\/mrow>x^2 + 6x + 5 = 0<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>+<\/span><\/span><\/span><\/span>6<\/span>x<\/span><\/span>+<\/span><\/span><\/span><\/span>5<\/span><\/span>=<\/span><\/span><\/span><\/span>0<\/span><\/span><\/span><\/span><\/span> by completing the square.<\/p>\n
        \n
      1. \n

        The equation is already in the correct form with a<\/mi>=<\/mo>1<\/mn><\/mrow>a = 1<\/annotation><\/semantics><\/math><\/span><\/span>a<\/span><\/span>=<\/span><\/span><\/span><\/span>1<\/span><\/span><\/span><\/span><\/span>.<\/p>\n<\/li>\n
      2. \n

        Move the constant term to the other side: x<\/mi>2<\/mn><\/msup>+<\/mo>6<\/mn>x<\/mi>=<\/mo>\u2212<\/mo>5<\/mn><\/mrow>x^2 + 6x = -5<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>+<\/span><\/span><\/span><\/span>6<\/span>x<\/span><\/span>=<\/span><\/span><\/span><\/span>\u2212<\/span>5<\/span><\/span><\/span><\/span><\/span>.<\/p>\n<\/li>\n
      3. \n

        Take half of the coefficient of x<\/mi><\/mrow>x<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span><\/span><\/span><\/span> and square it: (<\/mo>6<\/mn>2<\/mn><\/mfrac>)<\/mo><\/mrow>2<\/mn><\/msup>=<\/mo>3<\/mn>2<\/mn><\/msup>=<\/mo>9<\/mn><\/mrow>left(frac{6}{2}right)^2 = 3^2 = 9<\/annotation><\/semantics><\/math><\/span><\/span>(<\/span><\/span><\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>6<\/span><\/span><\/span><\/span><\/span>&ZeroWidthSpace;<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>)<\/span><\/span><\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span><\/span><\/span>3<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span><\/span><\/span>9<\/span><\/span><\/span><\/span><\/span>.<\/p>\n<\/li>\n
      4. \n

        Add this value to both sides of the equation: x<\/mi>2<\/mn><\/msup>+<\/mo>6<\/mn>x<\/mi>+<\/mo>9<\/mn>=<\/mo>\u2212<\/mo>5<\/mn>+<\/mo>9<\/mn><\/mrow>x^2 + 6x + 9 = -5 + 9<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>+<\/span><\/span><\/span><\/span>6<\/span>x<\/span><\/span>+<\/span><\/span><\/span><\/span>9<\/span><\/span>=<\/span><\/span><\/span><\/span>\u2212<\/span>5<\/span><\/span>+<\/span><\/span><\/span><\/span>9<\/span><\/span><\/span><\/span><\/span>.<\/p>\n<\/li>\n
      5. \n

        Rewrite the left side as a perfect square: (<\/mo>x<\/mi>+<\/mo>3<\/mn>)<\/mo><\/mrow>2<\/mn><\/msup>=<\/mo>4<\/mn><\/mrow>left(x + 3right)^2 = 4<\/annotation><\/semantics><\/math><\/span><\/span>(<\/span>x<\/span><\/span>+<\/span><\/span>3<\/span>)<\/span><\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span><\/span><\/span>4<\/span><\/span><\/span><\/span><\/span>.<\/p>\n<\/li>\n
      6. \n

        Solve for x<\/mi><\/mrow>x<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span><\/span><\/span><\/span>: x<\/mi>+<\/mo>3<\/mn>=<\/mo>\u00b1<\/mo>2<\/mn><\/mrow>x + 3 = pm2<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span>+<\/span><\/span><\/span><\/span>3<\/span><\/span>=<\/span><\/span><\/span><\/span>\u00b1<\/span>2<\/span><\/span><\/span><\/span><\/span>, so x<\/mi>=<\/mo>\u2212<\/mo>3<\/mn>\u00b1<\/mo>2<\/mn><\/mrow>x = -3 pm 2<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span>=<\/span><\/span><\/span><\/span>\u2212<\/span>3<\/span><\/span>\u00b1<\/span><\/span><\/span><\/span>2<\/span><\/span><\/span><\/span><\/span>. This gives us two solutions: x<\/mi>=<\/mo>\u2212<\/mo>1<\/mn><\/mrow>x = -1<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span>=<\/span><\/span><\/span><\/span>\u2212<\/span>1<\/span><\/span><\/span><\/span><\/span> and x<\/mi>=<\/mo>\u2212<\/mo>5<\/mn><\/mrow>x = -5<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span>=<\/span><\/span><\/span><\/span>\u2212<\/span>5<\/span><\/span><\/span><\/span><\/span>.<\/p>\n<\/li>\n<\/ol>\n\n\n\n

        How to Use the Formula to Solve Quadratic Equations<\/h3>\n\n\n\n
          \n
        1. \n

          Write down the quadratic equation in its standard form<\/strong>: a<\/mi>x<\/mi>2<\/mn><\/msup>+<\/mo>b<\/mi>x<\/mi>+<\/mo>c<\/mi>=<\/mo>0<\/mn><\/mrow>ax^2 + bx + c = 0<\/annotation><\/semantics><\/math><\/span><\/span>a<\/span>x<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>+<\/span><\/span><\/span><\/span>b<\/span>x<\/span><\/span>+<\/span><\/span><\/span><\/span>c<\/span><\/span>=<\/span><\/span><\/span><\/span>0<\/span><\/span><\/span><\/span><\/span>, where a<\/mi><\/mrow>a<\/annotation><\/semantics><\/math><\/span><\/span>a<\/span><\/span><\/span><\/span><\/span>, b<\/mi><\/mrow>b<\/annotation><\/semantics><\/math><\/span><\/span>b<\/span><\/span><\/span><\/span><\/span>, and c<\/mi><\/mrow>c<\/annotation><\/semantics><\/math><\/span><\/span>c<\/span><\/span><\/span><\/span><\/span> are coefficients and a<\/mi>\u2260<\/mo>0<\/mn><\/mrow>a neq 0<\/annotation><\/semantics><\/math><\/span><\/span>a<\/span><\/span><\/span>\ue020<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span><\/span><\/span><\/span>0<\/span><\/span><\/span><\/span><\/span>.<\/p>\n<\/li>\n
        2. \n

          Identify the coefficients<\/strong>: Determine the values of a<\/mi><\/mrow>a<\/annotation><\/semantics><\/math><\/span><\/span>a<\/span><\/span><\/span><\/span><\/span>, b<\/mi><\/mrow>b<\/annotation><\/semantics><\/math><\/span><\/span>b<\/span><\/span><\/span><\/span><\/span>, and c<\/mi><\/mrow>c<\/annotation><\/semantics><\/math><\/span><\/span>c<\/span><\/span><\/span><\/span><\/span> from the equation.<\/p>\n<\/li>\n
        3. \n

          Use the quadratic formula<\/strong>: Substitute the values of a<\/mi><\/mrow>a<\/annotation><\/semantics><\/math><\/span><\/span>a<\/span><\/span><\/span><\/span><\/span>, b<\/mi><\/mrow>b<\/annotation><\/semantics><\/math><\/span><\/span>b<\/span><\/span><\/span><\/span><\/span>, and c<\/mi><\/mrow>c<\/annotation><\/semantics><\/math><\/span><\/span>c<\/span><\/span><\/span><\/span><\/span> into the quadratic formula, which is x<\/mi>=<\/mo>\u2212<\/mo>b<\/mi>\u00b1<\/mo>b<\/mi>2<\/mn><\/msup>\u2212<\/mo>4<\/mn>a<\/mi>c<\/mi><\/mrow><\/msqrt><\/mrow>2<\/mn>a<\/mi><\/mrow><\/mfrac><\/mrow>x = frac{-b pm sqrt{b^2 – 4ac}}{2a}<\/annotation><\/semantics><\/math><\/span><\/span>x<\/span><\/span>=<\/span><\/span><\/span><\/span><\/span><\/span>2<\/span>a<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u2212<\/span>b<\/span>\u00b1<\/span><\/span>b<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u2212<\/span>4<\/span>a<\/span>c<\/span><\/span><\/span><\/span><\/path><\/svg><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>.<\/p>\n<\/li>\n
        4. \n

          Calculate the discriminant<\/strong>: The discriminant, D<\/mi><\/mrow>D<\/annotation><\/semantics><\/math><\/span><\/span>D<\/span><\/span><\/span><\/span><\/span>, is given by D<\/mi>=<\/mo>b<\/mi>2<\/mn><\/msup>\u2212<\/mo>4<\/mn>a<\/mi>c<\/mi><\/mrow>D = b^2 – 4ac<\/annotation><\/semantics><\/math><\/span><\/span>D<\/span><\/span>=<\/span><\/span><\/span><\/span>b<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u2212<\/span><\/span><\/span><\/span>4<\/span>a<\/span>c<\/span><\/span><\/span><\/span><\/span>. Calculate this value first as it will tell you how many real roots the equation has:<\/p>\n