imhimanshu wrote:

If ac≠0, do graphsy=ax2+b and y=cx2+dintersect?

(1) a=−c

(2) b>d

Hi,

Can anyone show me the underlying concept in this question.

Help will be appreciated.

Thanks

H

If ac\neq{0}, do graphs y=ax^2+b and y=cx^2+d intersect?Notice that graphs of

y=ax^2+b and

y=cx^2+d are parabolas.

Algebraic approach:(1)

a = -c. Given:

y_1= ax^2 + b and

y_2=-ax^2 + d. Now, if these two parabolas cross, then for some

x,

ax^2+b=-ax^2 + d should be true, which means that equation

2ax^2+(b-d)=0 must have a solution, some real root(s), or in other words discriminant of this quadratic equation must be

\geq0. So, the question becomes: is

discriminant=0-8a(b-d)\geq0? Or, is

discriminant=-8a(b-d)\geq0? Now can we determine whether this is true? We know nothing about

a,

b, and

d, hence no. Not sufficient.

(2)

b>d. The same steps: if

y_1= ax^2 + b and

y_2= cx^2 + d cross, then for some

x,

ax^2 +b=cx^2+d should be true, which means that equation

(a-c)x^2+(b-d)=0 must have a solution or in other words discriminant of this quadratic equation must be

\geq0. So, the question becomes: is

discriminant=0-4(a-c)(b-d)\geq0? Or, is

discriminant=-4(a-c)(b-d)\geq0? Now can we determine whether this is true? We know that

b-d>0 but what about

a-c? Hence no. Not sufficient.

(1)+(2) We have that

a=-c and

b>d, so

y_1= ax^2 + b and

y_2=-ax^2 + d. The same steps as above:

2ax^2+(b-d)=0 and the same question remains: is

discriminant=-8a(b-d)\geq0 true?

b-d>0 but what about

a? Not sufficient.

Answer: E.

Else consider two cases. First case:

y=-x^2+1 and

y=x^2+0 (upward and downward parabolas). Notice that these parabolas satisfy both statements and they cross each other;

Second case:

y=x^2+1 and

y=-x^2+0 (also upward and downward parabolas). Notice that these parabolas satisfy both statements and they do not cross each other.

Answer: E.

Also discussed here:

m24-12-formatting-error-73363.htmlHope it helps.

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