Question

Find all values of a such that a x^2 + 3x + 5 = 0 has two real roots.

Answer 1

Answer:

a < 9/20

Step-by-step explanation:

Our equation is the following: ax^2 + 3x + 5

a = unknown

b = 3

c = 5

We are concerned with the discriminant. Namely, b^2 - 4ac must be greater than 0 if we want to real roots

Lets plug in the values:

9 - 4a(5) > 0

9 - 20a > 0

9 > 20a

9/20 > a

a < 9/20

Answer 2

In a quadratic equation given as,

ax² + bx + c = 0

Discriminant of the equation → (b² - 4ac)

For the roots of the equation,

• Roots will be real if (b² - 4ac > 0)
• Roots will be complex if (b² - 4ac < 0)
• Only one root if b² - 4ac = 0

Given quadratic equation in the question is,

ax² + 3x + 5 = 0

Coefficient of x² = a

Coefficient of x = b = 3

Constant term = c = 5

If the roots of this equation are real,

b² - 4ac > 0

(3)² - 4a(5) > 0

9 - 20a > 0

20a < 9

a < $\frac{9}{20}$

Therefore, for all values of 'a' less than $\frac{9}{20}$, equation will have two real roots.

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