IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

gauravverma

how 2 solve the eqn

2 2

y =4x-3 and y=x

the intersection point

malay

int he first equation, substitute y=x
x=3x-4
giving x=2 hence y=2

kavincoolzz

well is the second eqn y^2=x or y=x^2??

hit_ur_heart

Dear gaurav , as such solving 4th degree equation isnt in the JEE syllabi , if it is given in the question, then there must be some special case involved , and as per expectation it turns out to be so :- now see
y ^2 = 4x -3 && y = x ^2

Substituting y = X^2 in first equatin we get :- x^4 -4x + 3=0
consider a function f(x) = x^4 -4x + 3.
f``(x) = 4*x^3 - 4 => the point of extremum is x = 1, as it is only solution on real line.
now see, f ``(x) = 12*x^3 >=0 and at x = 1 it is greater than 0
=> only extremum is point of minima
=> f(1) = 0 is the minimum value which is equal to 0.
=> the only solution is 1, because at all other points value will be more than this as f(1) is minima.
since it is 4 degree equation this implies its other 2 roots are complex and other two coinciding , i.e., x=1

Another fact that can be derived from is here is that since equation for the intersection of two curves has only one distinct solution that means 2 curve touch each other , they have same common tanget

This is the complete explanation, normally i do not provide such big explanation but in this case i tht it need to be explained

Cheers

rashmi_jain

hiiiiiiiiiiiiii

if ever you'll get a 4 degree equation in jee to solve,then it will be an easy one and you'll need to apply hit n trial method.

it can be clearly seen that x=1 satisfies the eqn.

always try using x=-2,-1,0,1,2

any one will definitely satisfy any four degree eqn given in jee.

but if u want the proper way ,then hit_ur_heart has explained the right way correctly.

but just a little correction,in this case f"(x)=12x^2 but not 12x^3

pavneet

he rashmi pls solve my question of ellipse

(pavneet)

thank you

Ask & Discuss Questions with Community & Experts