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How can you find solutions to a quadratic equation? - Answers

There are a number of different ways. The one way which always works is using the quadratic formula.So, the solutions of quadratic equation of the form ax2 + bx + c = 0 arex = [-b ± sqrt(b2 - 4ac)]/(2a)If b2 - 4ac, which is called the discriminant, is less than 0 then there is no real square root and so no real solution: if a > 0 the graph of the quadratic is either entirely above the x-axis and if a < 0 it is below. [The square roots do exist in the complex field but the fact that you asked this question indicates that you have not yet reached that level.]There are other methods such as completing the square which is, in fact, the same as the above but you go through a lot more steps before getting to the same point!Still another method is factorisation. This method is fine as long as you can easily work out two linear factors for the quadratic. A lot of high school quadratics that you will need to solve will be open to this approach but real life is not as simple!



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How can you find solutions to a quadratic equation? - Answers

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

There are a number of different ways. The one way which always works is using the quadratic formula.So, the solutions of quadratic equation of the form ax2 + bx + c = 0 arex = [-b ± sqrt(b2 - 4ac)]/(2a)If b2 - 4ac, which is called the discriminant, is less than 0 then there is no real square root and so no real solution: if a > 0 the graph of the quadratic is either entirely above the x-axis and if a < 0 it is below. [The square roots do exist in the complex field but the fact that you asked this question indicates that you have not yet reached that level.]There are other methods such as completing the square which is, in fact, the same as the above but you go through a lot more steps before getting to the same point!Still another method is factorisation. This method is fine as long as you can easily work out two linear factors for the quadratic. A lot of high school quadratics that you will need to solve will be open to this approach but real life is not as simple!



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

How can you find solutions to a quadratic equation? - Answers

There are a number of different ways. The one way which always works is using the quadratic formula.So, the solutions of quadratic equation of the form ax2 + bx + c = 0 arex = [-b ± sqrt(b2 - 4ac)]/(2a)If b2 - 4ac, which is called the discriminant, is less than 0 then there is no real square root and so no real solution: if a > 0 the graph of the quadratic is either entirely above the x-axis and if a < 0 it is below. [The square roots do exist in the complex field but the fact that you asked this question indicates that you have not yet reached that level.]There are other methods such as completing the square which is, in fact, the same as the above but you go through a lot more steps before getting to the same point!Still another method is factorisation. This method is fine as long as you can easily work out two linear factors for the quadratic. A lot of high school quadratics that you will need to solve will be open to this approach but real life is not as simple!

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      There are a number of different ways. The one way which always works is using the quadratic formula.So, the solutions of quadratic equation of the form ax2 + bx + c = 0 arex = [-b ± sqrt(b2 - 4ac)]/(2a)If b2 - 4ac, which is called the discriminant, is less than 0 then there is no real square root and so no real solution: if a > 0 the graph of the quadratic is either entirely above the x-axis and if a < 0 it is below. [The square roots do exist in the complex field but the fact that you asked this question indicates that you have not yet reached that level.]There are other methods such as completing the square which is, in fact, the same as the above but you go through a lot more steps before getting to the same point!Still another method is factorisation. This method is fine as long as you can easily work out two linear factors for the quadratic. A lot of high school quadratics that you will need to solve will be open to this approach but real life is not as simple!
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