Calculator UseThis online calculator is a quadratic equation solver that will solve a second-order polynomial equation such as ax2 + bx + c = 0 for x, where a ≠ 0, using the quadratic formula. Show
The calculator solution will show work using the quadratic formula to solve the entered equation for real and complex roots. Calculator determines whether the discriminant \( (b^2 - 4ac) \) is less than, greater than or equal to 0. When \( b^2 - 4ac = 0 \) there is one real root. When \( b^2 - 4ac > 0 \) there are two real roots. When \( b^2 - 4ac < 0 \) there are two complex roots. Quadratic Formula:The quadratic formula \( x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a } \) is used to solve quadratic equations where a ≠ 0 (polynomials with an order of 2) \( ax^2 + bx + c = 0 \) Examples using the quadratic formulaExample 1: Find the Solution for \( x^2 + -8x + 5 = 0 \), where a = 1, b = -8 and c = 5, using the Quadratic Formula. \( x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a } \) \( x = \dfrac{ -(-8) \pm \sqrt{(-8)^2 - 4(1)(5)}}{ 2(1) } \) \( x = \dfrac{ 8 \pm \sqrt{64 - 20}}{ 2 } \) \( x = \dfrac{ 8 \pm \sqrt{44}}{ 2 } \) The discriminant \( b^2 - 4ac > 0 \) so, there are two real roots. Simplify the Radical: \( x = \dfrac{ 8 \pm 2\sqrt{11}\, }{ 2 } \) \( x = \dfrac{ 8 }{ 2 } \pm \dfrac{2\sqrt{11}\, }{ 2 } \) Simplify fractions and/or signs: \( x = 4 \pm \sqrt{11}\, \) which becomes \( x = 7.31662 \) \( x = 0.683375 \) Example 2: Find the Solution for \( 5x^2 + 20x + 32 = 0 \), where a = 5, b = 20 and c = 32, using the Quadratic Formula. \( x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{ 2a } \) \( x = \dfrac{ -20 \pm \sqrt{20^2 - 4(5)(32)}}{ 2(5) } \) \( x = \dfrac{ -20 \pm \sqrt{400 - 640}}{ 10 } \) \( x = \dfrac{ -20 \pm \sqrt{-240}}{ 10 } \) The discriminant \( b^2 - 4ac < 0 \) so, there are two complex roots. Simplify the Radical: \( x = \dfrac{ -20 \pm 4\sqrt{15}\, i}{ 10 } \) \( x = \dfrac{ -20 }{ 10 } \pm \dfrac{4\sqrt{15}\, i}{ 10 } \) Simplify fractions and/or signs: \( x = -2 \pm \dfrac{ 2\sqrt{15}\, i}{ 5 } \) which becomes \( x = -2 + 1.54919 \, i \) \( x = -2 - 1.54919 \, i \) calculator updated to include full solution for real and complex roots How to Solve Quadratic Equations using the Quadratic FormulaThere are times when we are stuck solving a quadratic equation of the form a{x^2} + bx + c = 0 because the trinomial on the left side can’t be factored out easily. It doesn’t mean that the quadratic equation has no solution. At this point, we need to call upon the straightforward approach of the quadratic formula to find the solutions of the quadratic equation or put simply, determine the values of x that can satisfy the equation. In order use the quadratic formula, the quadratic equation that we are solving must be converted into the “standard form”, otherwise, all subsequent steps will not work. The goal is to transform the quadratic equation such that the quadratic expression is isolated on one side of the equation while the opposite side only contains the number zero, 0. Take a look at the diagram below.  In this convenient format, the numerical values of a, b, and c are easily identified! Upon knowing those values, we can now substitute them into the quadratic formula then solve for the values of x.
Slow down if you need to. Be careful with every step while simplifying the expressions. This is where common mistakes usually happen because students tend to “relax” which results to errors that could have been prevented, such as in the addition, subtraction, multiplication and/or division of real numbers. Examples of How to Solve Quadratic Equations by the Quadratic FormulaExample 1: Solve the quadratic equation below using the Quadratic Formula. By inspection, it’s obvious that the quadratic equation is in the standard form since the right side is just zero while the rest of the terms stay on the left side. In other words, we have something like this This is great! What we need to do is simply identify the values of a, b, and c then substitute into the quadratic formula. That’s it! Make it a habit to always check the solved values of x back into the original equation to verify. Example 2: Solve the quadratic equation below using the Quadratic Formula. This quadratic equation is absolutely not in the form that we want because the right side is NOT zero. I need to eliminate that 7 on the right side by subtracting both sides by 7. That takes care of our problem. After doing so, solve for x as usual. The final answers are {x_1} = 1 and {x_2} = - {2 \over 3}. Example 3: Solve the quadratic equation below using the Quadratic Formula. This quadratic equation looks like a “mess”. I have variable x‘s and constants on both sides of the equation. If we are faced with something like this, always stick to what we know. Yes, it’s all about the Standard Form. We have to force the right side to be equal to zero. We can do just that in two steps. I will first subtract both sides by 5x, and followed by the addition of 8. Values we need: a = - 1, b = - \,8, and c = 2 Example 4: Solve the quadratic equation below using the Quadratic Formula. Well, if you think that Example 3 is a “mess” then this must be even “messier”. However, you’ll soon realize that they are really very similar. We first need to perform some cleanup by converting this quadratic equation into standard form. Sounds familiar? Trust me, this problem is not as bad as it looks, as long as we know what to do. Just to remind you, we want something like this Therefore, we must do whatever it takes to make the right side of the equation equal to zero. Since we have three terms on the right side, it follows that three steps are required to make it zero. The solution below starts by adding both sides by 3{x^2}, followed by subtraction of 3x, and finally the addition of 5. Done! After making the right side equal to zero, the values of a, b, and c are easy to identify. Plug those values into the quadratic formula, and simplify to get the final answers! Example 5: Solve the quadratic equation below using the Quadratic Formula. First, we need to rewrite the given quadratic equation in Standard Form, a{x^2} + bx + c = 0.
After getting the correct standard form in the previous step, it’s now time to plug the values of a, b, and c into the quadratic formula to solve for x.
a = 1, b = - \,4, and c = - \,14
You might also be interested in: Solving Quadratic Equations by Square Root Method How do you solve an equation using the quadratic formula?The quadratic formula helps us solve any quadratic equation. First, we bring the equation to the form ax²+bx+c=0, where a, b, and c are coefficients. Then, we plug these coefficients in the formula: (-b±√(b²-4ac))/(2a) . See examples of using the formula to solve a variety of equations.
What are the 5 examples of quadratic equation?Examples of the standard form of a quadratic equation (ax² + bx + c = 0) include:. 6x² + 11x - 35 = 0.. 2x² - 4x - 2 = 0.. -4x² - 7x +12 = 0.. 20x² -15x - 10 = 0.. x² -x - 3 = 0.. 5x² - 2x - 9 = 0.. 3x² + 4x + 2 = 0.. -x² +6x + 18 = 0.. What are the 4 steps we used to solve using a quadratic formula?The four methods of solving a quadratic equation are factoring, using the square roots, completing the square and the quadratic formula.
What are the 3 quadratic formulas?There are three commonly-used forms of quadratics:. Standard Form: y = a x 2 + b x + c y=ax^2+bx+c y=ax2+bx+c.. Factored Form: y = a ( x − r 1 ) ( x − r 2 ) y=a(x-r_1)(x-r_2) y=a(x−r1)(x−r2). Vertex Form: y = a ( x − h ) 2 + k y=a(x-h)^2+k y=a(x−h)2+k.. |
Solve each equation using the quadratic formula
Related Posts
Copyright © 2024 moicapnhap Inc.
