Find the value of p and q such that
2x^2 - 4x + 3= 2 (x-p)^2 + q

PLEASE ANSWER!!!

The values of p and q such that 2·x² - 4·x + 3 = 2·(x - p)² + q, are;

p = 1, q = 1

Step-by-step explanation:

The given equation are;

2·x² - 4·x + 3 = 2·(x - p)² + q

By expanding the right hand side of the equation, we have;

2·x² - 4·x + 3 = 2·(x - p)² + q = 2·x² - 4·p·x + 2·p² + q

2·x² - 4·x + 3  = 2·x² - 4·p·x + 2·p² + q

Comparing the coefficient of x in both equations, gives;

2·x² + (- 4)·x + 3  = 2·x² + (- 4·p)·x + 2·p² + q

-4 = -4·p

∴ p = -4/(-4) = 1

p = 1

Comparing only the constant terms (terms not multiplied by "x" or "x²"), we get;

3 = 2·p² + q

Substituting the value for "p" obtained above (p = 1 ) in the previous equation, we get;

3 = 2 × 1² + q  = 2 + q

∴ q = 3 - 2 = 1

q = 1.

898.1666667

step-by-step explanation:

5389/6 = 898.1666667

answer: translate the circles so they share a common center point and dilate circle y by a scale factor of 3.