Question

Find values of a and b that make the following equality into identity:

Answer 1

To turn an equality into an identity, you would find values for 'a' and 'b' that work for all values of 'x'. The process involves equating coefficients and solving the resulting equations. Without specific equations, we must rely on general methods, such as using the quadratic formula when 'c' is known.

Step-by-step explanation:

Finding Values of a and b

To make the equality into an identity, you'll typically want to find values of 'a' and 'b' that satisfy the given equations for all values of 'x'. It seems like part of your question is missing, as we would need the specific equations to give a precise answer. However, the process involves equating coefficients of the same powers of 'x' from both sides of the equation and solving the resulting system of equations.

For example, if you have an equation of the form ax2 + bx + c = 0 and you want to make it an identity, you can find 'a' and 'b' by applying the quadratic formula when 'c' is given.

In the case of a quadratic equation ax2 + bx + c = 0, you can determine 'a' and 'b' by solving:

-b ± √(b2 - 4ac) / (2a)

Which simplifies down when given specific constants, as shown by the example where a = 3, b = 13, and c = -10. This yields two possible solutions for 'x' which can verify the values of 'a' and 'b'.

If an equation already has set constants like a = 1.00, b = 10.0, and c = -200, then the equation is already defined, and the values are given.

Answer 2

1.
`a=b=(3)/(4)`

2.
`a=-3,\ b=3`

Step-by-step explanation:

1. Given:

`(3x)/((x-2)(3x+2))=(a)/(x-2)+(b)/(3x+2)`

To find a and b, add two fractions in right side:

`(a)/(x-2)+(b)/(3x+2)\\ \\=(a(3x+2)+b(x-2))/((x-2)(3x+2))\\ \\=(3ax+2a+bx-2b)/((x-2)(3x+2))\\ \\=((3a+b)x+(2a-2b))/((x-2)(3x+2))`

So,

`(3x)/((x-2)(3x+2))=((3a+b)x+(2a-2b))/((x-2)(3x+2))`

Two fractions with same denominators are equal when they have equal numerators, so

`3x=(3a+b)x+(2a-2b)`

Equate coefficients:

`At\ x:\ \ 3=3a+b\\ \\At \ 1: 0=2a-2b`

From the second equation:

`a=b`

Substitute it into the first equation:

`3b+b=3\\ \\4b=3\\ \\b=(3)/(4)`

Hence,

`(3x)/((x-2)(3x+2))=((3)/(4))/(x-2)+((3)/(4))/(3x+2)`

2. Given:

`(3)/(x^2-5x+6)=(a)/(x-2)+(b)/(x-3)`

Note that
`(x-2)(x-3)=x^2-3x-2x+6=x^2-5x+6`

To find a and b, add two fractions in right side:

`(a)/(x-2)+(b)/(x-3)\\ \\=(a(x-3)+b(x-2))/((x-2)(x-3))\\ \\=(ax-3a+bx-2b)/((x-2)(x-3))\\ \\=((a+b)x+(-3a-2b))/((x-2)(x-3))`

So,

`(3)/((x-2)(x-3))=((a+b)x+(-3a-2b))/((x-2)(x-3))`

Two fractions with same denominators are equal when they have equal numerators, so

`3=(a+b)x+(-3a-2b)`

Equate coefficients:

`At\ x:\ \ 0=a+b\\ \\At \ 1: 3=-3a-2b`

From the first equation:

`a=-b`

Substitute it into the second equation:

`-3(-b)-2b=3\\ \\3b-2b=3\\ \\b=3\\ \\a=-3`

Hence,

`(3)/(x^2-5x+6)=(-3)/(x-2)+(3)/(x-3)`