(a) Express \(\frac{8x^2 + 8x + 9}{(x - 1)(2x + 3)^2}\) in partial fractions. (b) The coordinates of the centre and circumference of a circle are (-2, 5) an...

Answer Details

(a) \(\frac{8x^2 + 8x + 9}{(x - 1)(2x + 3)^2}≡\frac{A}{(x - 1)}+\frac{B}{2x + 3}+\frac{C}{(2x + 3)^2}\)

\(\frac{8x^2 + 8x + 9}{(x - 1)(2x + 3)^2}≡\frac{A(2x + 3)2+B(x - 1)(2x + 3)+C(x - 1)}{(x - 1)(2x + 3)^2}\)

\(8x^2+8x+9=A(2x+3)^2+B(x-1)(2x+3)+C(x-1)\)

Put \(x=1\)

\(8(1)^2+8(1)+9=A(2(1)+3)^2+B(1-1)(2(1)+3)+C(1-1)\)

⇒25=25A

=A=\(\frac{25}{25}=1\)

Put \(x=-\frac{3}{2}\)

\(8(-\frac{3}{2})^2+8(-\frac{3}{2})+9=A(2(-\frac{3}{2})+3)^2+B(-\frac{3}{2}-1)(2(-\frac{3}{2})+3)+C(-\frac{3}{2}-1)\)

⇒15=-2.5C

\(=C=-\frac{15}{2.5}=-6\)

Since \(8x^2+8x+9=A(2x+3)^2+B(x-1)(2x+3)+C(x-1)\)

\(⇒8x^2+8x+9=A(4x^2+12x+9)+B(2x^2+x-3)+C(x-1)\)

\(=8x^2+8x+9=4Ax^2+12A+9A+2Bx^2+Bx-3B+Cx-C\)

\(=8x^2+8x+9=4Ax^2+2Bx^2+12Ax+Bx+Cx+9A-3B-C\)

\(=8x^2+8x+9=(4A+2B)x^2+(12A+B+C)x+9A-3B-C\)

By comparing the coefficient of \(x, 8=12A+B+c\)

=8=12(1)+B-6

=8=12+B-6

=8=6+B

=8-6=B

=\(\therefore \frac{8x^2+8x+9}{(x-1)(2x+3)^2}=\frac{1}{x-1}+\frac{2}{2x+3}-\frac{6}{(2x+3)^2}\)

(b) Equation of a circle =\((x - a)^2 + (y - b)^2 = r^2\)

Where "a" and "b" are the coordinate of the center and "r" is the radius

2πr = 6π (given)

∴ r = 3 units

=\( (x - (-2))^2 + (y - 5)^2 = 3^2\)

= \((x + 2)^2 + (y - 5)^2 = 9\)

= \(x^2 + 4x + 4 + y^2 - 10y + 25 = 9\)

= \(x^2 + y^2 + 4x - 10y + 29 - 9 = 0\)

∴ \(x^2 + y^2 + 4x - 10y + 20 = 0\)