Question

If alpha beta are roots of equation X square - 2 X + 2 equal to zero then Alpha square plus beta square is

Answer 1

Given:

$\bold{ \alpha \: and \: \beta \: are \: the \: roots \: of \: equation}$

$\star \bold{f(x) = {x}^{2} - 2x + 2 = 0}$

To Find :

$\star \bold{ { \alpha }^{2} + { \beta }^{2} }$

Here :

a = 1 , b=-2 and c= 2

As we know that :

$\star \: \alpha + \beta = \frac{ - b}{a} = \frac{ - ( - 2)}{1} = 2 \\ \\ \star \: \alpha \beta = \frac{c}{a} = \frac{2}{1} = 2$

Now :

$\implies \: { \alpha }^{2} + { \beta }^{2} = ( { \alpha + \beta )}^{2} - 2 \alpha \beta \\ \\ \implies \: { \alpha }^{2} + { \beta }^{2} = ( {2)}^{2} - 2(2) \\ \\ \implies \: { \alpha }^{2} + { \beta }^{2} = 4 - 4 \\ \\ \implies \: { \alpha }^{2} + { \beta }^{2} = 0$

Hence ,The value is 0

Answer 2

QUESTION :-

=> If alpha beta are roots of equation X square - 2 X + 2 equal to zero then Alpha square plus beta square is.

SOLUTION :-

=> Let, α and β be roots of equation.

=> f(x) = x²-2x + 2 = 0

=> On comparing with ax²+bx+c then we get,

=> • a = 1 • b = -2. • c = 2

=> .°. Sum of zeroes = α + β

=> -b/a = -(-2)/1 = 2.

=> .°. Product of zeroes = αβ

=> c/a = 2/1 = 2.

Now,

=> We have to find out, α² + β²,

=> .°. α² + β² = (α + β)² - 2αβ

=> α²+β² = (2)²-2(2)

=> α²+β² = 4 - 4

=> .°. α²+β² = 0

Therefore, the value of α²+β² is 0.