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Determine the number of real solutions to the quadratic equation 5a2−6a+10=0.

Select the correct answer below:
0 real solutions
1 real solution
2 real solutions

User Tavares
by
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1 Answer

5 votes

Answer:

There are no real solutions to the quadratic equation 5a² - 6a + 10 = 0.

Explanation:

We can determine the number of real solutions to a quadratic equation according to the value of its determinant, which is written with the capital letter D.

If pa² + qa + r = 0 and p > 1.

Thus, D = q² - 4pr.

If D > 0, then the equation has 2 real solutions.

If D = 0, then the equation has 1 real solution.

If D < 0, then the equation has 0 real solutions.

We can deduce D from the equation 5a² - 6a + 10 = 0 by writing the following:

p = 5

q = -6

r = 10

D = q² - 4pr

D = (-6)² - 4(5)(10)

D = 36 - 200

D = -164

D < 0

Since the determinant is negative, the equation 5a² - 6a + 10 = 0 has zero real solutions.

I hope this helps! Sorry if my English didn't really help with having a clearer explanation.

User Kamyar Gilak
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