Question

Use graph of the function f(x)=x2 to find how the number of roots of the equation depends on the value of b.

a)x^2=x−b

If b < ANSWER, the equation has 2 roots.
If b = ANSWER, the equation has 1 root.
If b > ANSWER, the equation has no roots.

b) x^2=bx−1


If b is on the interval ( , ) ∪ ( , ), the equation has two roots.
If b equals to , , the equation has one root.
If b is on the interval ( , ), the equation has no roots.

Answer 1

a)

• 2 roots: b < 1/4
• 1 root: b = 1/4
• 0 roots: b > 1/4

b)

• 2 roots: b ∈ (-∞, -2) ∪ (2, ∞)
• 1 root: b ∈ {-2, 2}
• 0 roots: b ∈ (-2, 2)

Explanation:

a) Writing the equation in standard form gives ...

x^2 -x +b = 0

Then the discriminant* is ...

d = (-1)^2 -4(1)(b) = 1-4b

Solving for b, we have ...

b = (1 -d)/4

When d > 0, b < 1/4 and there will be 2 roots.

When d = 0, b = 1/4 and there will be 1 root.

When d < 0, b > 1/4 and there will be no real roots.

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b) We can work the second problem in similar fashion. The standard form of the given equation is ...

x^2 -bx +1 = 0

The discriminant of this is ...

d = (-b)^2 -4(1)(1) = b^2 -4

d = (b +2)(b -2)

When d > 0, b ∈ (-∞, -2) ∪ (2, ∞) and there are 2 roots.

When d = 0, b ∈ {-2, 2}, and there is one root.

When d < 0, b ∈ (-2, 2), and there are no real roots.

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* The discriminant of the equation ...

ax^2 + bx + c = 0

is ...

d = b^2 -4ac

When it is positive, there are two real roots; when it is zero, there is one real root; when it is negative, there are no real roots.