Question

The zeros of x^2-3x+2 are alpha and beta.Then the value of alpha^2+beta^2​

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore \alpha^{2}+\beta^{2}=5}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline { \underline\bold{Given : }} \\ \implies x \in( {x}^{2} - 3x + 2 = 0) \\ \\ \underline { \underline\bold{To \: Find: }} \\ \implies {\alpha }^{2} + { \beta }^{2} = ?$

• According to given question :

$\bold{Using \: middle \: term \: spliting : } \\ \implies {x}^{2} - 3x + 2 = 0 \\ \\ \implies {x}^{2} - 2x - x + 2 = 0 \\ \\ \implies x(x - 2) - 1(x - 2) = 0 \\ \\ \implies (x - 2)(x - 1) = 0 \\ \\ \bold{\implies x = 2 \: and \: 1} \\ \\ \bold{ \alpha = 2} \\ \bold{ \beta = 1} \\ \\ \bold{For \: finding \: values : } \\ \implies { \alpha }^{2} + { \beta }^{2} = {(\alpha + \beta )}^{2} - 2 \alpha \beta \\ \\ \implies { \alpha }^{2} + { \beta }^{2} = {(2 + 1)}^{2} - 2 \times 2 \times 1 \\ \\ \implies { \alpha }^{2} + { \beta }^{2} = {3}^{2} - 4 \\ \\ \implies { \alpha }^{2} + { \beta }^{2} = 9 - 4 \\ \\ \implies \bold{{ \alpha }^{2} + { \beta }^{2} = 5}$

Answer 2

SOLUTION:-

Given:

The zeros of x² -3x +2 are alpha & beta.

To find:

The value of alpha² + beta².

Explanation:

We have,

f(x)= x² -3x +2 & these equation is compare with Ax² + Bx +C, we get;

•A = 1

•B= -3

•C= 2

Therefore,

We know that, sum of the roots;

$\alpha + \beta = - \frac{b}{a} = - \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

So,

$= > \alpha + \beta = \frac{ - ( - 3 )}{1} \\ \\ = > \alpha + \beta = \frac{3}{1} \\ \\ = > \alpha + \beta = 3$

&

Product of the roots,

$= > \alpha \times \beta = \frac{c}{a} = \frac{constant \: term}{coefficient \: of \: {x}^{2} }$

So,

=) Alpha× beta= 2/1

=) alpha × beta= 2

Now,

Symmetric functions:

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

So,

Putting the values;

$= > (3) {}^{2} - 2 \times 2 \\ \\ = > 9 - 4 \\ \\ = > 5$

Therefore,

(alpha)² + (beta)²= 5

RajMaur.