Question

a, B are the zeroes of x²+x+1
find a+B+ ab​

Answer 1

Answer:

$p(x) = x {}^{2} + x + 1 \\ x {}^{2} + (1 + 1)x + 1 \\ x {}^{2} + x + x + 1 \\ x(x +1 ) + 1(x + 1) \\ ( x + 1)(x + 1) \\ x + 1 = 0 \\ x = - 1 \\ x + 1 = 0 \\ x = - 1 \\ let \: \alpha \: and \: \beta \: be \: the \: zeos \: of \: p(x) \\ sum \: of \: zeros \: = \alpha + \beta = - 1 + ( - 1) \\ \frac{ - b}{a} = \frac{ - 2}{1} \\ \alpha + \beta = - 2 \: .........(i) \\ product \: of \: zeros \: = \alpha \beta \\ \frac{c}{a} = - 1 \times ( - 1) \\ \alpha \beta = 1 \: ......(ii) \\ from \: i \: and \: \: ii \\ \alpha + \beta + \alpha \beta = - 2 + 1 = - 1$

Answer 2

CORRECT QUESTION:

If α , β are the zeroes of the quadratic equation

f(x) = x² + x + 1 . Then find α + β + αβ

ANSWER:

METHOD 1:

Firstly finding α , β by quadratic equation formula

x = -b±√b²-4ac/2a

→ x² + x + 1 = 0

Here , a = 1 , b = 1 , c = 1

➵ x = - 1 ±√1² - 4(1)(1)/2(1)

➵ x = -1±√1 - 4/2

➵ x = -1±√-3/2

➵ x = - 1 + √3(√-1)/2 or -1 - √3(-√1)/2

Using complex numbers

➳ √-1 = i

➳ -1 = i²

➵ x = -1+√3 i/2 or -1 - √3i/2

Now,

α = √3i - 1 /2 , β = -√3i- 1/2

Finding : α + β + αβ

★ - 1 + √3i/2 + (- 1 -√3i/2 ) + (√3i-1/2)(-√3i - 1/2)

★ - 2/2 + (√3i - 1)(-√3i - 1)/4

★ - 1 + (-3i² - √3i + √3i + 1)/4

★ - 1 + 4/4

★ - 1 + 1

★ 0

METHOD 2 :

Using

Sum of roots (α + β) = -b/a

Product of roots (αβ) = c/a

Finding : α + β + αβ

♦ - b/a + c/a

♦ -1/1 + 1/1

♦ - 1 + 1

♦ 0

Hence, α + β + αβ = 0