Determine the value(s) of m so that the quadratic equation mx2+6x = −m has no real roots. (i.e. no solution)
help please !
no real roots means the discriminant is negative.
mx^2 + 6m + m = 0
b^2-4ac = 36-4m^2
so 9-m^2 < 0
To find the values of m for which the quadratic equation mx^2 + 6x = -m has no real roots, we can analyze the discriminant of the quadratic equation. The discriminant determines the nature of the roots.
The general form of a quadratic equation is ax^2 + bx + c = 0, where a, b, and c are constants. In this case, the given equation is mx^2 + 6x = -m, which can be rewritten as mx^2 + 6x + m = 0.
The discriminant (denoted as Δ) of a quadratic equation is given by the formula Δ = b^2 - 4ac. If Δ is greater than zero, there are two distinct real roots. If Δ is equal to zero, there is one real root (repeated root). If Δ is less than zero, there are no real roots.
In our case, comparing the equation to the general form ax^2 + bx + c, we have a = m, b = 6, and c = m.
Substituting these values into the formula, we get:
Δ = (6)^2 - 4(m)(m)
Δ = 36 - 4m^2
Δ = 4(9 - m^2)
We want to find the values of m for which Δ < 0, indicating no real roots.
Since Δ is a product of 4 and (9 - m^2), if Δ < 0, then (9 - m^2) must be positive. Hence:
9 - m^2 > 0
To solve this inequality, we can rearrange it as follows:
m^2 - 9 < 0
(m - 3)(m + 3) < 0
Next, we construct a sign chart by considering the inequality when m - 3 = 0 and m + 3 = 0:
-3 3
|---|---|
m - +
It shows that for m < -3 or m > 3, the inequality (m - 3)(m + 3) < 0 holds true.
Therefore, the values of m for which the quadratic equation mx^2 + 6x = -m has no real roots are m < -3 or m > 3.