# Find the parameter a that belongs to R if parabola y = (a + 1)*x^2 + a*x + 3 and the line y = x + 1 have two distinct points in common.

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### 3 Answers

To find the parameree a of y = (a+1)x^2+ax+3 and y = x+1 has distinct point of interception.

At the point of interception. ordinates are same both curves:

(a+1)x^2+ax+3 = x+1. Or

(a+1)x^2+(a-1)x+2 = 0

The discriminant (a-1)^2 -4(a+1)*2> 0 for distinct points of intersection.

a^2 -2a+1 - 8a-8 > 0

a^2 -10a -7 > 0

a1 = {10 +sqr(10^2+4*7)}/2 = 5+ 4sqrt2

a2 = 5 - 4sqr2.

Therefore a should not be in the interval ( 10 -4 sqrt2 , 10+4sqrt2 ) in order that the incepts are different.

At the point of intersection of y = (a + 1)*x^2 + a*x + 3 and y = x + 1, the y- values are equal. Therefore equating them we get:

(a + 1)*x^2 + a*x + 3 = x + 1

=> (a + 1)*x^2 + a*x - x + 3 - 1 = 0

=> (a + 1)*x^2 + (a -1 ) *x + 2 = 0

Now according to the initial condition there are two distinct points in common, therefore (a-1)^2 - 4*(a+1)*2 > 0

=> a^2 - 2a +1 - 8a - 8 > 0

For a^2 - 10a - 7 =0

the roots are [10 + sqrt( 100 + 28)] /2 and [10 - sqrt ( 100 + 28)]/2

or 5 + 4 sqrt 2 and 5 - sqrt 2.

Therefore as a^2 - 2a +1 - 8a - 8 should be greater than 0, a should lie either below 5 - 4 sqrt 2 or above 5 + 4 sqrt 2.

**a can be either below 5 - 4 sqrt 2 or above 5 + 4 sqrt 2.**

The line y=x+1 and the parabola have 2 distinct points only if the equation (a+1)*x^2 + a*x + 3 = x+1 have 2 distinct roots, meaning that the equation (a+1)*x^2 + x*(a-1) + 2=0 has to have the discriminant delta > 0.

delta = (a-1)^2 - 8*(a+1)=a^2-2*a+1-8*a-8=a^2-10*a-7>0

The roots of the equation a^2-10*a-7=0 are

a1=[10+(100+28)]/2=5+4*sqrt(2)

a2=5-4*sqrt(2)

We'll apply the rule: between the roots we have the opposite sign of the coefficient of x^2, that means minus, and outside the roots, the same sign, meaning plus.

So the expression is positive on the following intervals:

(-infinity,5-4*(2)^1/2)U(5+4*(2)^1/2, infinity)