What is the point of intersection of y = 4x^2 + 3 and y = 6 - 2x^2

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The point of intersection of the parabolas y = 4x^2 + 3 and y = 6 - 2x^2 has to be determined. At the point of intersection, the values of the x and y coordinates of both the parabolas is the same.

Solve 4x^2 + 3 = 6 - 2x^2

=> 6x^2 = 3

=> x^2 = 1/2

=> x = `sqrt (1/2)` and x = `-sqrt(1/2)`

For x = `sqrt (1/2)` , y = 5 and for x = `-sqrt(1/2)` , also y = 5

**The points of intersection of the two parabolas is `(sqrt(1/2), 5)` and `(-sqrt(1/2), 5)` **

You need to solve the system of simultaneous equations to find the point of intersection of the given curves such that:

`{(y = 4x^2 + 3),(y = 6 - 2x^2):}`

You need to use substitute `4x^2 + 3` for y in the second equation such that:

` 4x^2 + 3 = 6 - 2x^2 => 4x^2 + 3 - 6 + 2x^2 = 0 => 6x^2 - 3 = 0`

`6x^2 = 3 => x^2 = 3/6 => x^2 = 1/2 => x_(1,2) = +-sqrt2/2`

You need to substitute `+-sqrt2/2` for x in the first or the second equation such that:

`4/2 + 3 = y_1 => y_1 = 5 = y_2`

**Hence, evaluating the solutions to the system that represent the points of intersection of the given curves yields `(sqrt2/2 ; 5) and (-sqrt2/2 ; 5). ` **

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