Question

If a + b = 4 and ab = -12, find (1) a - b (ii) 2 - 2​

Answer 1

Step-by-step explanation:

ANSWER:-

Before we go ahead, identities to be used are

${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy > > > (1)$

${(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy > > (2)$

Using Above identities:-

We will attempt the question as follows.

Now , First using identity (2),

${(a + b)}^{2} = {(4)}^{2} = 16$

We can write it as

$16 = {a}^{2} + {b}^{2} + 2( - 12)$

${a}^{2} + {b}^{2} = 24 + 16 = 40$

Using the above,

Now using identity (2),

${(a - b)}^{2} = 40 - 2( - 12)$

• Here I placed the above we found directly,

${(a - b)}^{2} = 64$

$1)a - b = 8 \: \longrightarrow \: ( \sqrt{64} = 8)$

Note:-

• Second equation ie 2-2 is wrong as it does not apply any identity

Answer 2

(1)

Given that :

• a + b = 4
• ab = -12

We know that,

(a + b)² - (a - b)² = 4ab

Using this identity,

→ (4)² - (a - b)² = 4(-12)

→ 16 - (a - b)² = -48

→ -(a - b)² = -16 - 48

→ -(a - b)² = -64

→ (a - b)² = 64

→ a - b = √64

→ a - b = 8

(ii)

2 - 2 = 0

Any number subtracted from itself except infinity is always 0.