2x^2 - 20x + c ≡ a(x - b)^2 + 3b Work out the value of c.?

2x² - 20x + c ≡ a(x - b)²+ 3b
2x² - 20x + c ≡ a(x² - 2bx + b²) + 3b
2x² - 20x + c ≡ ax² - 2bx + (b² + 3b)

Compare the x² terms on the both sides :
a = 2

Compare the x terms on the both sides :
-20 = -2b
b = 10

Compare the constant terms on the both sides :
c = b² + 3b
c = 10² + 3*10
c = 130

2x² - 20x + c ≡ a(x - b)² + 3b

We need to determine the values of a, b and c so that the expression on the left is identical to the expression on the right for ANY value of x.

2x² - 20x + c ≡ a(x² - 2bx + b²) + 3b

so, 2x² - 20x + c ≡ ax² - 2abx + ab² + 3b

Then, equating coefficients of x², x and constants we have:

a = 2

2ab = 20.....so, 4b = 20...i.e. b = 5

ab² + 3b = c....so, 2(25) + 15 = c => 65

Therefore, 2x² - 20x + 65 ≡ 2(x - 5)² + 15

Again, as we have an identity, if we choose any value of x, the left side will give exactly the same value as the right side.

:)>

" _"
.._
.._
means " is identical to "
The given is called the identity.
We say that the expression on the
left is identical to that on the right.
Given : 2x^2-20x+c=a(x-b)^2+3b identically
=>2x^2-20x+c=a(x^2-2bx+b^2)+3b
=>2x^2-20x+c=ax^2-2abx+ab^2+3b
In this case, we can compare the coefficients
of the different like terms:
a=2
-2ab=-20
c=ab^2+3b
Solving the system of a, b,c get
c=2(5^2)+3(5)=65