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Unit 14: Lesson 6
- The quadratic formula
- Understanding the quadratic formula
- Worked example: quadratic formula (example 2)
- Worked example: quadratic formula (negative coefficients)
- Using the quadratic formula: number of solutions
- Quadratic formula review
- Discriminant review
Quadratic formula
- (Choice A) x = 3 , − 1 2 x =3,-\frac{1}{2} x = 3 , − 2 1 x, equals, 3, comma, minus, start fraction, 1, divided by, 2, end fraction A x = 3 , − 1 2 x =3,-\frac{1}{2} x = 3 , − 2 1 x, equals, 3, comma, minus, start fraction, 1, divided by, 2, end fraction
- (Choice B) x = 5 ± 57 16 x =\dfrac{5 \pm \sqrt{57}}{16} x = 1 6 5 ± 5 7 x, equals, start fraction, 5, plus minus, square root of, 57, end square root, divided by, 16, end fraction B x = 5 ± 57 16 x =\dfrac{5 \pm \sqrt{57}}{16} x = 1 6 5 ± 5 7 x, equals, start fraction, 5, plus minus, square root of, 57, end square root, divided by, 16, end fraction
- (Choice C) x = 1 ± 17 − 4 x =\dfrac{1 \pm \sqrt{17}}{-4} x = − 4 1 ± 1 7 x, equals, start fraction, 1, plus minus, square root of, 17, end square root, divided by, minus, 4, end fraction C x = 1 ± 17 − 4 x =\dfrac{1 \pm \sqrt{17}}{-4} x = − 4 1 ± 1 7 x, equals, start fraction, 1, plus minus, square root of, 17, end square root, divided by, minus, 4, end fraction
- (Choice D) x = − 4 ± 34 3 x =\dfrac{-4 \pm \sqrt{34}}{3} x = 3 − 4 ± 3 4 x, equals, start fraction, minus, 4, plus minus, square root of, 34, end square root, divided by, 3, end fraction D x = − 4 ± 34 3 x =\dfrac{-4 \pm \sqrt{34}}{3} x = 3 − 4 ± 3 4 x, equals, start fraction, minus, 4, plus minus, square root of, 34, end square root, divided by, 3, end fraction
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Section 2.5 : Quadratic Equations - Part I
For problems 1 – 7 solve the quadratic equation by factoring.
- \({u^2} - 5u - 14 = 0\) Solution
- \({x^2} + 15x = - 50\) Solution
- \({y^2} = 11y - 28\) Solution
- \(19x = 7 - 6{x^2}\) Solution
- \(6{w^2} - w = 5\) Solution
- \({z^2} - 16z + 61 = 2z - 20\) Solution
- \(12{x^2} = 25x\) Solution
For problems 8 & 9 use factoring to solve the equation.
- \({x^4} - 2{x^3} - 3{x^2} = 0\) Solution
- \({t^5} = 9{t^3}\) Solution
For problems 10 – 12 use factoring to solve the equation.
- \(\displaystyle \frac{{{w^2} - 10}}{{w + 2}} + w - 4 = w - 3\) Solution
- \(\displaystyle \frac{{4z}}{{z + 1}} + \frac{5}{z} = \frac{{6z + 5}}{{{z^2} + z}}\) Solution
- \(\displaystyle x + 1 = \frac{{2x - 7}}{{x + 5}} - \frac{{5x + 8}}{{x + 5}}\) Solution
For problems 13 – 16 use the Square Root Property to solve the equation.
- \(9{u^2} - 16 = 0\) Solution
- \({x^2} + 15 = 0\) Solution
- \({\left( {z - 2} \right)^2} - 36 = 0\) Solution
- \({\left( {6t + 1} \right)^2} + 3 = 0\) Solution
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How to Solve Quadratic Equations
Last Updated: February 10, 2023 References Approved
This article was co-authored by David Jia . 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. There are 9 references cited in this article, which can be found at the bottom of the page. wikiHow marks an article as reader-approved once it receives enough positive feedback. This article has 15 testimonials from our readers, earning it our reader-approved status. This article has been viewed 1,299,454 times.
A quadratic equation is a polynomial equation in a single variable where the highest exponent of the variable is 2. [1] X Research source 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.
Factoring the Equation
- 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] X Research source
- 3x = -1 ..... by subtracting
- 3x/3 = -1/3 ..... by dividing
- x = -1/3 ..... simplified
- 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.
- So, both solutions do "check" separately, and both are verified as working and correct for two different solutions.
Using the Quadratic Formula
- 4x 2 - 5x - 13 = x 2 -5
- 4x 2 - x 2 - 5x - 13 +5 = 0
- 3x 2 - 5x - 8 = 0
- {-b +/-√ (b 2 - 4ac)}/2
- {-(-5) +/-√ ((-5) 2 - 4(3)(-8))}/2(3) =
- {-(-5) +/-√ ((-5) 2 - (-96))}/2(3)
- {-(-5) +/-√ ((-5) 2 - (-96))}/2(3) =
- {5 +/-√(25 + 96)}/6
- {5 +/-√(121)}/6
- (5 + 11)/6 = 16/6
- (5-11)/6 = -6/6
- x = (-1, 8/3)
Completing the Square
- 2x 2 - 9 = 12x =
- In this equation, the a term is 2, the b term is -12, and the c term is -9.
- 2x 2 - 12x - 9 = 0
- 2x 2 - 12x = 9
- 2x 2 /2 - 12x/2 = 9/2 =
- x 2 - 6x = 9/2
- -6/2 = -3 =
- (-3) 2 = 9 =
- x 2 - 6x + 9 = 9/2 + 9
- x = 3 + 3(√6)/2
- x = 3 - 3(√6)/2)
Practice Problems and Answers
Expert Q&A
- 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: [14] X Research source ⧼thumbs_response⧽ Helpful 0 Not Helpful 0
- 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. ⧼thumbs_response⧽ Helpful 0 Not Helpful 0
- If the "b" is an even number, the formula is : {-(b/2) +/- √(b/2)-ac}/a. ⧼thumbs_response⧽ Helpful 0 Not Helpful 0
You Might Also Like
- ↑ https://www.mathsisfun.com/definitions/quadratic-equation.html
- ↑ http://www.mathsisfun.com/algebra/factoring-quadratics.html
- ↑ https://www.mathportal.org/algebra/solving-system-of-linear-equations/elimination-method.php
- ↑ https://www.cuemath.com/algebra/quadratic-equations/
- ↑ https://www.purplemath.com/modules/solvquad4.htm
- ↑ http://www.purplemath.com/modules/quadform.htm
- ↑ https://uniskills.library.curtin.edu.au/numeracy/algebra/quadratic-equations/
- ↑ http://www.mathsisfun.com/algebra/completing-square.html
- ↑ http://www.umsl.edu/~defreeseca/intalg/ch7extra/quadmeth.htm
About This Article
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! Did this summary help you? Yes No
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Solving quadratic equations
In this GCSE Maths Edexcel study guide, you'll learn how to solve quadratic equations by factorising, using formulae and completing the square. Each method also provides information about the corresponding quadratic graph.
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Quadratic equations
A quadratic equation contains terms up to \(x^2\) . There are many ways to solve quadratics. All quadratic equations can be written in the form \(ax^2 + bx + c = 0\) where \(a\) , \(b\) and \(c\) are numbers ( \(a\) cannot be equal to 0, but \(b\) and \(c\) can be).
Here are some examples of quadratic equations in this form:
- \(2x^2 - 2x - 3 = 0\) . \(a = 2\) , \(b = -2\) and \(c = -3\)
- \(2x(x + 3) = 0\) . \(a = 2\) , \(b = 6\) and \(c = 0\) (in this example, the bracket can be expanded to \(2x^2 + 6x = 0\) )
- \((2x + 1)(x - 5) = 0\) . \(a = 2\) , \(b = -9\) and \(c = -5\) (this will expand to \(2x^2 - 9x - 5 = 0\) )
- \(x^2 + 2 = 4\) . \(a = 1\) , \(b = 0\) and \(c = -2\)
- \(3x^2 = 48\) . \(a = 3\) , \(b = 0\) and \(c = -48\) (in this example \(c = -48\) , but has been rearranged to the other side of the equation)
\(3x^2 = 48\) is an example of a quadratic equation that can be solved simply.
Divide both sides by 3
Square root both sides to isolate x
The positive and negative square roots of 16 are 4 and -4, as 4 × 4 = 16 and -4 × -4 = 16, so x = ± 4
Solving by factorising
If the product of two numbers is zero then one or both of the numbers must also be equal to zero. Therefore if \(ab = 0\) , then \(a = 0\) and/or \(b = 0\) .
If \((x + 1)(x + 2) = 0\) , then \(x + 1 = 0\) or \(x + 2 = 0\) , or both. Factorising quadratics will also be used to solve these equations.
Expanding the brackets \((x + 2)(x + 3)\) gives \(x^2 + 3x + 2x + 6\) , which simplifies to \(x^2 + 5x + 6\) . Factorising is the reverse process of expanding brackets, so factorising \(x^2 + 5x + 6\) gives \((x + 2)(x + 3)\) .
Solve \(x(x + 3) = 0\) .
The product of \(x\) and \(x + 3\) is 0, so \(x = 0\) or \(x + 3 = 0\) , or both.
\[\begin{array}{rcl} x + 3 & = & 0 \\ -3 && -3 \\ x & = & -3 \end{array}\]
\(x = 0\) or \(x = -3\) .
Solve \((x + 1)(x - 5) = 0\) .
The product of \(x + 1\) and \(x - 5\) is 0, so one or both brackets must also be equal to 0.
\[\begin{array}{rcl} x + 1 & = & 0 \\ -1 && -1 \\ x & = & -1 \end{array}\]
\[\begin{array}{rcl} x - 5 & = & 0 \\ + 5 && + 5 \\ x & = & 5 \end{array}\]
\(x = -1\) or \(x = 5\)
Solve \(x^2 + 7x + 12 = 0\) .
Begin by factorising the quadratic .
The quadratic will be in the form \((x + a)(x + b) = 0\) .
Find two numbers with a product of 12 and a sum of 7.
\(3 \times 4 = 12\) , and \(3 + 4 = 7\) , so \(a\) and \(b\) are equal to 3 and 4. This gives:
\[(x + 3)(x + 4) = 0\]
The product of \(x + 3\) and \(x + 4\) is 0, so \(x + 3 = 0\) or \(x + 4 = 0\) , or both.
\[\begin{array}{rcl} x + 3 & = & 0 \\ - 3 && - 3 \\ x & = & -3 \end{array}\]
\[\begin{array}{rcl} x + 4 & = & 0 \\ - 4 && - 4 \\ x & = & -4 \end{array}\]
\(x = -3\) or \(x = -4\)
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How to use solver on ti84 to solve quadratic equations
Use your TI 84 Calculator to find the solution to a quadratic equation or the factored form of a quadratic expression. For more TI 84
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TI 84 Plus CE Solving Quadratic Equations with the
Typing in the quadratic formula on a TI-84 CE to calculate the solutions to a quadratic equaiton. This is just one of many ways to do this.
A demonstration of the techniques for finding the zeros or roots of a quadratic function to solve a quadratic equation on the Ti84 graphing
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If you have a general quadratic equation like this: ax^2+bx+c=0 ax2 + bx + c = 0 Then the formula will help you find the roots of a quadratic equation, i.e. the values of x x where this equation is solved. The quadratic formula x=\dfrac {-b\pm\sqrt {b^2-4ac}} {2a} x = 2a−b ± b2 − 4ac It may look a little scary, but you'll get used to it quickly!
The quadratic formula Quadratic formula CCSS.Math: HSA.REI.B.4, HSA.REI.B.4b Google Classroom Solve. 6 + 2x^2 - 3x = 8x^2 6 + 2x2 −3x = 8x2 Choose 1 answer: x =3,-\frac {1} {2} x = 3,−21 A x =3,-\frac {1} {2} x = 3,−21 x =\dfrac {5 \pm \sqrt {57}} {16} x = 165 ± 57 B x =\dfrac {5 \pm \sqrt {57}} {16} x = 165 ± 57
How to Solve a Quadratic Equation Using the Quadratic Formula Example 9.3. 1 Solve 2 x 2 + 9 x − 5 = 0 by using the Quadratic Formula. Answer Example 9.3. 2 Solve 3 y 2 − 5 y + 2 = 0 by using the Quadratic Formula. Answer Example 9.3. 3 Solve 4 z 2 + 2 z − 6 = 0 by using the Quadratic Formula. Answer
Remember, the quadratic formula looks like this: ± √ ( Our coefficients: , , and Our equation after inserting the coefficients: ± √ ( 5 Use the order of operations to simplify the formula. Download Article Just do the math in the equation as you normally would.
Quadratic Formula Calculator Watch on Example (Click to try) 2 x 2 − 5 x − 3 = 0 About the quadratic formula Solve an equation of the form a x 2 + b x + c = 0 by using the quadratic formula: x = Quadratic Formula Video Lesson Solve with the Quadratic Formula Step-by-Step [1:29] Need more problem types? Try MathPapa Algebra Calculator Show Keypad
This video explains how to solve quadratic equations using the quadratic formula. My Website: https://www.video-tutor.net How To Use The Quadratic Formula To Solve Equations How to...
For problems 1 - 7 solve the quadratic equation by factoring. u2 −5u−14 = 0 u 2 − 5 u − 14 = 0 Solution x2 +15x =−50 x 2 + 15 x = − 50 Solution y2 = 11y−28 y 2 = 11 y − 28 Solution 19x = 7−6x2 19 x = 7 − 6 x 2 Solution 6w2 −w =5 6 w 2 − w = 5 Solution z2 −16z +61 = 2z −20 z 2 − 16 z + 61 = 2 z − 20 Solution 12x2 = 25x 12 x 2 = 25 x Solution
Solve by using the Quadratic Formula: x(x + 6) + 4 = 0. Solution: Our first step is to get the equation in standard form. Distribute to get the equation in standard form. This equation is now in standard form. Identify the values of a, b, c. Write the Quadratic Formula. Then substitute in the values of a, b, c. Simplify. Simplify the radical.
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. Method 1 Factoring the Equation 1
Learn about quadratic equations using our free math solver with step-by-step solutions. Skip to main content. Microsoft Math Solver. Solve Practice Download. ... Type a math problem. Type a math problem. Solve. Examples. x^2-3x=28. x ^ { 2 } - 5 x + 3 y = 20. x^2-10x+25=0. 2x^2+12x+40=0 ...
In algebra, a quadratic equation (from Latin quadratus 'square') is any equation that can be rearranged in standard form as where x represents an unknown value, and a, b, and c represent known numbers, where a ≠ 0. (If a = 0 and b ≠ 0 then the equation is linear, not quadratic.) The numbers a, b, and c are the coefficients of the equation ...
Use the quadratic formula to find the solutions. −b±√b2 −4(ac) 2a - b ± b 2 - 4 ( a c) 2 a Substitute the values a = 1 a = 1, b = 7 b = 7, and c = 10 c = 10 into the quadratic formula and solve for x x. −7±√72 −4 ⋅(1⋅10) 2⋅1 - 7 ± 7 2 - 4 ⋅ ( 1 ⋅ 10) 2 ⋅ 1 Simplify. Tap for more steps... x = −7±3 2 x = - 7 ± 3 2
This algebra video tutorial explains how to solve quadratic equations by factoring in addition to using the quadratic formula. This video contains plenty of examples and practice...
There are many ways to solve quadratics. All quadratic equations can be written in the form \ (ax^2 + bx + c = 0\) where \ (a\), \ (b\) and \ (c\) are numbers (\ (a\) cannot be equal to 0, but...
A demonstration of the techniques for finding the zeros or roots of a quadratic function to solve a quadratic equation on the Ti84 graphing Homework Support Solutions Solving math equations can be challenging, but it's also a great way to improve your problem-solving skills.