isn't it sppose to be : x^2 + (a+b)* x + ab ?Tryingmybest wrote:y=(x+a)(x+b)
= x^2 -(a+b)x +ab --- 1
This intesects y axis at (o,-6)
0= 36 -(a+b)6 + ab
Now we should know a + b to solve the equation which is provided by A
0= 36 +6 + ab
=> ab=-42
So 1 becomes x^2 +x -42 = 0
(X-7) (X+6) =0
So intesection points will be ( 7,0) and (-6,0)
So C
Tryingmybest wrote:Yeah ....thanks for pointing out.
But still the solution is C.
y=(x+a)(x+b)
= x^2 +(a+b)x +ab --- 1
This intesects y axis at (o,-6)
=>-6=ab
Now we should know a + b to solve the equation which is provided by A
So 1 becomes x^2 -x -6= 0
(X-3) (X+2) =0
So intesection points will be ( -2,0) and (3,0)
So C
It is perfectly Fine, often some curves intersect the axes at 2 points [e.g. parabola]bhumika.k.shah wrote:But is it fine if we are getting two values for x ? that is -2 and 3 ??
bhumika.k.shah wrote:Huh?????ajith wrote: [e.g. parabola]
bhumika.k.shah wrote:But is it fine if we are getting two values for x ? that is -2 and 3 ??
Question itself asks for two values of X "what two points does the graph"
The two points where the graph of y = (x + a)(x + b) intersect the x-axis are the roots of the quadratic equation x^2 + (a + b) x + a b = 0. Hence, if (a + b) and a b are known, it's done.gorsar wrote:In the xy-plane, at what two points does the graph of y=(x+a)(x+b) intersect
the x-axis?
1) a+b=-1
2) The graph intersects the y-axis at (0,-6)
OG answer is C.
Can anyone explaine why answer is C?
