Question

a^+b^+2b+4a+5=0,then the value of (a-b)/(a+b) is​

Answer 1

Answer:

Step-by-step explanation:

Given :

a² + b² + 2b + 4a + 5 = 0

To Find :

The value of : $\frac{a - b}{a + b}$

Solution :

a² + b² + 2b + 4a + 5 = 0

⇒ (a² + 4a + 4) + (b² + 2b + 1) = 0

⇒ [(a)² + 2(2)(a) + (2)²] + [(b)^2 + 2(1)(b) + (1)²]

⇒ (a + 2)² + (b + 1)² = 0

Note :

If sum of two squares are zero, then, the numbers MUST BE EQUAL TO ZERO.

⇒ (a + 2) = 0 ⇒ a = -2

⇒ (b + 1) = 0 ⇒ b = -1,.

Hence,

$\frac{a - b}{a + b} = \frac{-2 - ( -1)}{ -2 + (-1)}$

⇒ $\frac{-2 + 1}{-2 - 1} = \frac{-1}{-3} = \frac{1}{3}$

∴ $\frac{a - b}{a + b} = \frac{1}{3}$

Answer 2

$\huge \mathfrak \red{ANSWER :}$

undefined plz write your question clearly.

$\green{STEP-BY-STEP \: \: VERIFICATION:}$

$\red{ {a}^{2} + {b}^{2} + 2b + 4a + 5 = 0} \\ \red{= >} \green{ {a}^{2} + 4a + (b {}^{2} + 2b + 5) = 0} \\ \purple{= > }a = \frac{ - b \frac{ + }{} \sqrt{ {b}^{2} -4ac } }{ 2a } \\ = > \blue{a = \frac{ - 4 \frac{ + }{} \sqrt{ {4}^{2} - 4 \times 1 \times ( {b}^{2} + 2b + 5)} }{2 \times 1} } \\ \orange{= > } \pink{a = \frac{ - 4 \frac{ + }{} \sqrt{ {16} - ( 4{b}^{2} + 8b + 20)} }{2 } } \\ = > a = undfined \: since \: \: \sqrt{ - {?}^{} } \: \: is \: imaginary.$