Question

Why is the answer B? Plz explain

Answer 1

I wonder what the best way of explaining this is? You could graph the results for each choice. Or you could reason it out. 

The first thing you have to do it deal with y = 3. Draw a really crude graph like a set of axis. Make the y values go from -5 to + 5. x for the moment does not matter. 

Draw a horizontal line through y = 3 or going through the point (0,3). Now here's the catch and you're going to have to read it very carefully.

Condition One
If a>0 then the graph opens upward and b is going to have to be less than 3. That sentence is an absolute nightmare. Think carefully about what a>0 means. Make it 2 and draw a rough graph opening upward on the crude axis you have drawn. b is the y value of the minimum, so that minimum has to be less than 3.  Are there any point like that? The two points where a>0 are C and D. 

C
The lowest point that C will hit is 4.  That's not good enough. b = 4 is the lowest y value. C is not the answer

D 
The lowest point is 3. The graph will just touch y = 3. That's not good enough either. Touching y = 3 does not produce 2 roots. It produces just one.

Condition Two
a < 0 Here the graph opens downward. It means the a<0 and b>3. You need to look at A or B. Which point does that? A has a maximum below 3. It's no good. A is wrong.

So the answer must be B. a<0 and b>3. Right on 2 solutions.

Answer 2

Solution below assumes knowledge of the quadratic formula for the equation
$ax^2+bx+c=0$ where solution
x=$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
and where the expression $b^2-4ac$ is called the discriminant.

Given system of linear equations:
y=3 ..................................(1)
y=ax^2+b.........................(2)  
[ note: b here is the constant term, and is "c" in the quadratic formula]

Need to find a and b so that system has exactly two solutions, i.e. two distinct solutions, not one, not zero.

Substitute (1) in (20
3=ax^2+b
rearrange
ax^2+b-3=0 ........................(3)

Solve by the quadratic formula:
x=(-0 &pm; &radic; (0^2-4a(b-3))/2a
For x to have two distinct solutions, the discriminant must be greater than zero.
The discriminant is the expression inside the square-root sign, namely
0^2-4a(b-3) 
=-4a(b-3)

case A: a=-2,b=2, discriminant = -4(-2)(2-3)=-8 <0  (no real solution)
case B: a=-2,b=4, discriminant = -4(-2)(4-3)=+8 >0  (2 real distinct solutions)
case C: a= 2,b=4, discriminant = -4(2)(4-3)=-8 <0  (no real solution)
case D: a= 4,b=3, discriminant = -4(4)(3-3)= 0 =0  (two coincident solutions)

So from the previous calculations, the only case that gives two distinct real solutions is case B, which gives a discriminant >0.