a The quadratic equation x^2=d has no real solution if and only if d<0.
B
b The quadratic equation x^2=d has exactly one solution if and only if d=0.
C
c The quadratic equation x^2=d has two solutions if and only if d>0.
A
aExample Equation: 2x^2+8=0
B
bExample Equation: 2x^2=0
Solution: x=0
C
cExample Equation: 2x^2-8=0
Solutions: x=± 2
Practice makes perfect
a Let's first recall that a quadratic equation of the form x^2=d has no real solutions if and only if d<0. With this in mind, let's isolate x in the equation ax^2+c=0.
Note that we are told that a≠0, so we know that the right-hand side of the above equation is defined. Now, thinking back to what we said before, the equation x^2=- ca has no real solutions if and only if - ca is less than 0.
- c/a<0 ⇔ c/a>0
The quotient ca is greater than 0 when a and c are nonzero numbers with the same sign — both positive or both negative. Therefore, one of the infinitely many possible combinations of values is a=2 and c=8.
2x^2+8=0
Let's try to solve the above equation and verify that there are no real solutions.
Since there is no real number whose square is - 4, the above equation has no real solution. ✓
b Let's first recall that a quadratic equation of the form x^2=d has exactly one solution if and only if d=0. With this in mind, let's isolate x in the equation ax^2+c=0.
Note that we are told that a≠0, so the right-hand side of the above equation is defined. Based on our previous reasoning, the equation x^2=- ca has exactly one solution if and only if - ca is equal to 0.
- c/a=0 ⇔ c=0
Assuming a≠0, the equation ax^2+c=0 has exactly one solution if and only if c=0. Therefore, a can take any value other than 0. Let's consider, for example, a=2.
2x^2+0=0 ⇔ 2x^2=0
Let's solve the above equation and verify that it has exactly one solution.
Our equation has exactly one solution, which is x=0. ✓
c Let's first recall that a quadratic equation of the form x^2=d has two solutions if and only if d>0. With this in mind, let's isolate x in the equation ax^2+c=0.
Note that we are told that a≠0, so the right-hand side of the above equation is defined. According to what we said before, the equation x^2=- ca has two solutions if and only if - ca is greater than 0.
- c/a>0 ⇔ c/a<0
The quotient ca is less than 0 when a and c are nonzero numbers with opposite signs — one positive and one negative. Therefore, one of the infinitely many possible pairs of values is a=2 and c=- 8.
2x^2+(- 8)=0 ⇔ 2x^2-8=0
Let's solve the above equation and verify that it has two solutions.
