Question

please answer in nb
intelligent can only solve in nb
solve 1 question​

Answer 1

Given:

• 2x² -4x -1
• alpha and beta are zeroes of the polynomial

To find:

$The \: valu e \: of \: ( \alpha + \beta ), ( \alpha \beta ) \: and \: ( \alpha + \beta ) {}^{2}$

Solution:

Comparing 2x² -4x -1 with standard equation ax² +bx +c

We get,

a = 2

b = -4

c = -1

We know that

$Sum \: of \: zeroes \: = \frac{ - (coefficient \: of \: x)}{ \: \: \: coefficient \: of \: {x}^{2} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{ - b}{a}$

Here,$\:\alpha \: and \: \beta \: are \: zeroes \:$

So,

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

Also,

$Product \: of \: zeroes = \frac{constant \: term }{coefficient \: of \: {x}^{2} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{c}{a}$

$\large \implies \: ( \alpha \beta ) = \: \frac{ - 1}{ \: 2}$

We will use algebraic identity to find the value of (α²+ β²).

We know that

=> (a+b)² = a² + b² + 2ab

=> (a+b)² -2ab = a² + b²

Similarly,

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

Now we will substitute the values from above

$\implies \: { \alpha }^{2} + { \beta }^{2} = (2) {}^{2} - 2 \times ( \frac{ - 1}{2} ) \\ \\ \implies \: { \alpha }^{2} + { \beta }^{2} = 4 + \frac{ \cancel2} { \cancel2} \\ \\ \implies \: { \alpha }^{2} + { \beta }^{2} = 4$

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