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Algebra Level 2

How many real roots does the equation ( 4 x − 17 ) ( x 2 − 4 x + 13 ) ( x 2 + 6 x + 1 ) = 0 (4x-17)(x^2-4x+13)(x^2+6x+1) = 0 have?


The answer is 3.

Ethan Song by Ethan Song , 22, USA

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1 solution

Ethan Song
Jun 22, 2017

Because the equation is already factored, we can just find the discriminant of each quadratic to determine the number of real roots.

  • 4 x − 17 = 0 4x-17 = 0 gives us 1 real solution, 17 4 \frac{17}{4} , since all linear equations give 1 solution.

  • x 2 − 4 x + 13 = 0 x^2-4x+13 = 0 gives us 0 real solutions, since the discriminant of the equation is 16 − 52 = n e g a t i v e 16 - 52 = negative .

  • x 2 + 6 x + 1 = 0 x^2+6x+1 = 0 gives us 2 real solutions, since the discriminant of the equation is 36 − 4 = p o s i t i v e 36 - 4 = positive .

Thus, there are 3 3 real solutions to the equation.

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