Question

Perimeter of a rectangle is 64 cm and its area is 84 sq cm. Which one is the second degree equation relating to the given condition? *

Options
_________

a) x² + 32x + 84 = 0
b) x² - 32x + 84 = 0
c) x² + 32x -84 = 0​

Answer 1

Answer:

c) x² + 32x -84 = 0​. Pls mark me as brainliest.

Step-by-step explanation:

2(l+b) = 64

l+b = 32

l = b - 32

Let breadth be x

l = x - 32

lb = 84

x(x - 32) = 84

$x^2 - 32x = 84\\x^2 - 32x - 84 = 0$

Answer 2

Step-by-step explanation:

ANSWER

2(l+b) = 64

2(l+b) = 64l+b = 32

2(l+b) = 64l+b = 32l = b - 32

2(l+b) = 64l+b = 32l = b - 32Let breadth be x

2(l+b) = 64l+b = 32l = b - 32Let breadth be xl = x - 32

2(l+b) = 64l+b = 32l = b - 32Let breadth be xl = x - 32lb = 84

2(l+b) = 64l+b = 32l = b - 32Let breadth be xl = x - 32lb = 84x(x - 32) = 84

2(l+b) = 64l+b = 32l = b - 32Let breadth be xl = x - 32lb = 84x(x - 32) = 84\begin{gathered}x^2 - 32x = 84\\x^2 - 32x - 84 = 0\end{gathered}

{x}^{2} - 32x = 84

{x}^{2} - 32x - 84 = 0