math dq 7-1 - You plug the solutions a and b into(x-a(x-b =...

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The discriminant b^2 - 4ac, tells us how many real number solutions the equation ax^2 + bx + c= 0 has. When D = negative, has two non real imaginary number solutions. When D = 0, it has only one solution, it is a real number b/c zero is the perfect square and can be factored as a square When D = positive it has two different real number solutions How do you find a quadratic equation if you are only given the solution?
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Unformatted text preview: You plug the solutions a and b into: (x-a)(x-b) = 0 If you multiply a quadratic equation by a contant number, you do not change the solutions. So x^2 = 3 and 2x^2 = 6 have the same solutions. Find a quadratic equation that has roots x = 2 and x = 5: (x-2)(x-5) = 0. Find a quadratic equation that has two roots at x = 1: (x-1)(x-1) = 0....
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