Prove that - A quadratic equation cannot have more than 2 roots Please don't copy from any other websites and answer correct

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Q: 3. What do we know about the graph of a quadratic that has imaginary roots? (zeros with i)

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Q: how to divide the parabola using long division (x4-16)/(x-2) and x4 is x to the 4th power

A: The given division,   x4-16x-2

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A: Given equation is               2x2-7x=15      or. 2x2-7x-15=0

Q: | Using the rational root theorem for the polynominl 5z–2²+2-12, what are the values of esting?

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Q: or False. When solving a polynomial of degree 4, we will always obtain 4 real solutions.

A: When solving a polynomial of degree 4, Let  f(x) = (x-2)2*(x-3)2

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Q: Solve. 3 v+2-851 If all real numbers are solutions, click on "All reals". If there is no solution,…

A: Recall: x≤a ⇒ -a≤x≤a

Q: FInd all real solutions.   Square root of (3 - square root of [x + 5] ) = 2

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Q: Prove that the cubic root of 5 is irrational.  Please show all steps. Thank you!!!

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Q: What is the nature of the roots of the quadratic equation if the value of its discriminant is zero?

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A: discriminant number of solution D>0 2 real numbers D=0 1 real number D<0 no real numbers

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Q: Solve without

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Roots

The main dissimilarity between real and complex roots is that the real roots are obtained as real numbers, in contrast with the complex roots, which are indicated as imaginary numbers. The standard format for depicting any complex number is in the form of a + bi.