Question

If a, ß are the roots of x2 – 2x + 1 = 0 then find alpha^2+beta^2

Answer 1

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Answer 2

Answer:

$\large\boxed{\sf{2}}$

Step-by-step explanation:

Given a quadratic eqn such that,

${x}^{2} - 2x + 1 = 0$

Also, we know that general eqn for a quadratic equation,

• $a {x}^{2} + bx + c = 0$

On comparing the coefficients, we have,

• a = 1
• b = -2
• c = 1

Also, it's given that the zeroes are,

• $\alpha$
• $\beta$

Now, we know that, sum of roots,

$= > \alpha + \beta = - \dfrac{b}{a} \\ \\ = > \alpha + \beta = - (\frac{ - 2}{1} ) \\ \\ = > \alpha + \beta = - ( - 2) \\ \\ = > \alpha + \beta = 2$

And, Product of roots,

$= > \alpha \beta = \dfrac{c}{a} \\ \\ = > \alpha \beta = \dfrac{1}{1} \\ \\ = > \alpha \beta = 1$

Now, to find the value of,

$= > { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta ) }^{2} - 2 \alpha \beta$

Substituting the values, we get,

$= > { \alpha }^{2} + { \beta }^{2} = {2}^{2} - 2 \times 1 \\ \\ = > { \alpha }^{2} + { \beta }^{2} = 4 - 2 \\ \\ = > { \alpha }^{2} + { \beta }^{2} = 2$

Hence, the required value is 2.