Question #fa20c

Question

1457 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-24T23:50:59

See below.

The identities

$I_{1} \to a {(x - 1)}^{2} + b (2 x - 1) (x - 1) + c (x + 3) \equiv (5 - x)$ $I_1->a (x - 1)^2 + b (2 x - 1) (x - 1) + c (x + 3) equiv (5 - x)$

$I_{2} \to a {(x - 2)}^{2} + b {(x - 1)}^{2} + c \equiv (3 x^{2} - 8)$ $I_2->a (x - 2)^2 + b (x - 1)^2 + c equiv (3 x^2 - 8)$

are obtained after expanding and grouping coefficients given

$I_1->{(a+b+3c=5),(2a+3b=1),(a+2b=0):}$

and solving we obtain $a=4,b=-2,c=1$

so

$I_1->4 (x - 1)^2 -2 (2 x - 1) (x - 1) + (x + 3) equiv (5 - x)$

analogously we have

$I_2->-3 (x - 2)^2 +6 (x - 1)^2 -2 equiv (3 x^2 - 8)$