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How does an incomplete quadratic equation differ from a complete quadratic equation? - Answers

A complete quadratic equation is expressed in the form ax2+bx+c=0 (for example, 1x2-5x+4=0), and the formula to solve it is x1,2=(-b±√[b2-4ac])/2a {using the numbers from the example: x1,2=(5±√[25-16])/2 x1,2=(5±√[9])/2 x1,2=(5±3)/2 So the answers will be 4 ([5+3]/2) and 1 ([5-3]/2).} An incomplete quadratic equation can be expressed in two forms: ax2+bx=0 and ax2+c=0. The first form is solved by taking x (in some cases, x multiplied by a number that both a and b can be divided by, like in the next example) out of the two numbers, making the equation x(ax+b)=0. Then, either the outcome of the brackets (in the next example, the outcome of the brackets is x+1, so if x+1=0, then x=-1) or the x multiplying them needs to be zero in order for the equation to be correct, for example: 5x2+5x=0 5x(x+1)=0 x1=0 x2=(-1) Or by taking out just x: 5x2+5x=0 x(5x+5)=0 x1=0 x2=(-1) The second form is solved like a regular equation, for example: 2x2-162=0 2x2=162 x2=81 x=±9



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How does an incomplete quadratic equation differ from a complete quadratic equation? - Answers

https://math.answers.com/math-and-arithmetic/How_does_an_incomplete_quadratic_equation_differ_from_a_complete_quadratic_equation

A complete quadratic equation is expressed in the form ax2+bx+c=0 (for example, 1x2-5x+4=0), and the formula to solve it is x1,2=(-b±√[b2-4ac])/2a {using the numbers from the example: x1,2=(5±√[25-16])/2 x1,2=(5±√[9])/2 x1,2=(5±3)/2 So the answers will be 4 ([5+3]/2) and 1 ([5-3]/2).} An incomplete quadratic equation can be expressed in two forms: ax2+bx=0 and ax2+c=0. The first form is solved by taking x (in some cases, x multiplied by a number that both a and b can be divided by, like in the next example) out of the two numbers, making the equation x(ax+b)=0. Then, either the outcome of the brackets (in the next example, the outcome of the brackets is x+1, so if x+1=0, then x=-1) or the x multiplying them needs to be zero in order for the equation to be correct, for example: 5x2+5x=0 5x(x+1)=0 x1=0 x2=(-1) Or by taking out just x: 5x2+5x=0 x(5x+5)=0 x1=0 x2=(-1) The second form is solved like a regular equation, for example: 2x2-162=0 2x2=162 x2=81 x=±9



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https://math.answers.com/math-and-arithmetic/How_does_an_incomplete_quadratic_equation_differ_from_a_complete_quadratic_equation

How does an incomplete quadratic equation differ from a complete quadratic equation? - Answers

A complete quadratic equation is expressed in the form ax2+bx+c=0 (for example, 1x2-5x+4=0), and the formula to solve it is x1,2=(-b±√[b2-4ac])/2a {using the numbers from the example: x1,2=(5±√[25-16])/2 x1,2=(5±√[9])/2 x1,2=(5±3)/2 So the answers will be 4 ([5+3]/2) and 1 ([5-3]/2).} An incomplete quadratic equation can be expressed in two forms: ax2+bx=0 and ax2+c=0. The first form is solved by taking x (in some cases, x multiplied by a number that both a and b can be divided by, like in the next example) out of the two numbers, making the equation x(ax+b)=0. Then, either the outcome of the brackets (in the next example, the outcome of the brackets is x+1, so if x+1=0, then x=-1) or the x multiplying them needs to be zero in order for the equation to be correct, for example: 5x2+5x=0 5x(x+1)=0 x1=0 x2=(-1) Or by taking out just x: 5x2+5x=0 x(5x+5)=0 x1=0 x2=(-1) The second form is solved like a regular equation, for example: 2x2-162=0 2x2=162 x2=81 x=±9

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      A complete quadratic equation is expressed in the form ax2+bx+c=0 (for example, 1x2-5x+4=0), and the formula to solve it is x1,2=(-b±√[b2-4ac])/2a {using the numbers from the example: x1,2=(5±√[25-16])/2 x1,2=(5±√[9])/2 x1,2=(5±3)/2 So the answers will be 4 ([5+3]/2) and 1 ([5-3]/2).} An incomplete quadratic equation can be expressed in two forms: ax2+bx=0 and ax2+c=0. The first form is solved by taking x (in some cases, x multiplied by a number that both a and b can be divided by, like in the next example) out of the two numbers, making the equation x(ax+b)=0. Then, either the outcome of the brackets (in the next example, the outcome of the brackets is x+1, so if x+1=0, then x=-1) or the x multiplying them needs to be zero in order for the equation to be correct, for example: 5x2+5x=0 5x(x+1)=0 x1=0 x2=(-1) Or by taking out just x: 5x2+5x=0 x(5x+5)=0 x1=0 x2=(-1) The second form is solved like a regular equation, for example: 2x2-162=0 2x2=162 x2=81 x=±9
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