Discriminants and Inequalities?
Hey, just got confused with a question on AQA Core 1 Maths on the Specimen Paper.
Here's the question:
The number x satisfies the equation:
x^2 + mx + 16 = 0
where m is a constant,
Find the values of m for which this equation has:
(a) Equal roots
(b) 2 Roots
(c) No roots
I understand (a) but I don't understand b or c
For b I got m > +/- 8 and for c I got m < +/- 8
When the answers actually are:
(a) m = +/- 8
(b) m < -8 or m > 8
(c) -8 < m < 8
Answers & Comments
Verified answer
Given a quadratic equation of the form ax² + bx + c = 0, then the nature of the roots is given by the discriminant of the quadratic formula.
i.e. b² - 4ac
so, with a = 1, b = m and c = 16 we have:
m² - 64
There are three scenarios.
m² - 64 = 0.....equal roots
so, m² = 64
=> m = ±√64
i.e. m = ±8
Next, when m² - 64 > 0....two real roots
i.e. (m + 8)(m - 8) > 0
so, when m < -8 and when m > 8
Lastly, m² - 64 < 0....no real roots
i.e. (m + 8)(m - 8) < 0
so, -8 < m < 8
A sketch will help you see the regions better:
http://www.wolframalpha.com/input/?i=y+%3D+m%C2%B2...
The first part is where the curve crosses the m axis.
The second part is where the curve is above the axis.
The third part is where the curve is below the axis.
:)>
The general solution of the Equation is
x=-m/2 +- sqrt(m^2/4 -16)
One doble root when (m^2)/4=16, m=+-8
Two real roots when (m^2)/4-16>0, ImI>8
two imaginary roots when (m^2/4-16)<0, ImI<8