Question

4) Sendo P = y + 3 e Q = y - 3, calcule:
a) P^2
b) Q^2
c) P.Q ​

Answer 1

$P = y + 3\\Q = y - 3$

a)

$P^2 = (y + 3)^2\\\\(y + 3)^2 = y^2 + 2*3*y + 3^2 \\\\y^2 + 6y + 9\\\\\\\frac{-b \pm \sqrt{b^2 - 4ac} }{2a} \\\\\frac{-6 \pm \sqrt{6^2 - 4*9} }{2} = \frac{-6 \pm \sqrt{0} }{2} = \frac{-6 }{2} = -3$

b)

$Q^2 = (y - 3)^2\\\\(y - 3) = y^2 - 2*3*y + 3^2\\\\y^2 - 6y + 9\\\\\frac{-b \pm \sqrt{b^2 - 4ac} }{2a}\\\\\frac{-(-6) \pm \sqrt{(-6)^2 - 4*9} }{2} = \frac{6 \pm \sqrt{0} }{2} = \frac{6}{2} = 3$

c)

$P*Q = (y + 3)*(y - 3) = y^2 - 3^2 = y^2 - 9\\\\y^2 - 9 = 0\\y^2 = 9\\y = \pm \sqrt{9} = \pm 3$