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If b²-4ac > 0 and a non-perfect square, then...

User Ejaz Khan
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Final answer:

If b² - 4ac > 0 and a non-perfect square, a quadratic equation has two real and distinct solutions.

Step-by-step explanation:

A quadratic equation in the form ax² + bx + c = 0 can be solved using the quadratic formula:

x = (-b ± √(b² - 4ac))/(2a)

In this case, if b² - 4ac > 0 and a is a non-perfect square, then the quadratic equation has two real and distinct solutions.

For example, if we have an equation like x² + 4x + 3 = 0, where a = 1, b = 4, and c = 3, then we can calculate the solutions using the quadratic formula:

x = (-4 ± √(4² - 4(1)(3)))/(2(1))

x = (-4 ± √(16 - 12))/(2)

x = (-4 ± √4)/(2)

x = (-4 ± 2)/(2)

Therefore, the solutions are x = -3 and x = -1.

User Indya
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