Question

if alpha amd beta are the zeroes of the polynomial x2+7x+3 then the value of (alpha-beta) whole square is​

Answer 1

Step-by-step explanation:

Solution :

If α and β are the zeroes of the quadratic polynomial x² + 7x + 3

The value of (α - β)²

As we know that polynomial are compared with ax² + bx + c

a = 1

b = 7

c = 3

1st

Condition:

We now that sum of the zeroes;

2nd

Condition:

We know that product of the zeroes;

Now;

We know that formula of the (α - β)² :

⟹(α+β)

2

−4αβ

⟹(−7)

2

−4(3)

⟹−7×(−7)−4×3

⟹49−12

⟹37

Thus;

The value of (α - β)² = 37 .

Answer 2

$\underline{ \huge\bf\red{QuestiØn} :}$

$\sf if \: \alpha \: and \: \beta \: are \: the \: zeroes \: \\ \sf of \: the \: polynomial \: \mathcal{{x}^{2} + 7x + 3} \\ \sf then \: find \: the \: value \: of \: {( \alpha - \beta )}^{2} .$

$\underline{ \: \huge\bf\green{SolutiØn} \: :}$

As per question,

we have,

• sum of zeroes, ( α + β ) = 7/1 = 7

• product of zeroes, ( α•β ) = 3/1 = 3

Thus,

$\begin{aligned} \\ & \quad \ \ {( \alpha - \beta )}^{2} \\ \\ & = {( \alpha - \beta )}^{2} + 2 \alpha \beta - 2 \alpha \beta \\ \\ & = { \alpha }^{2} + { \beta }^{2} - 2 \alpha \beta + 2\alpha \beta - 2 \alpha \beta & \\ \\ & = ( { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta ) - 4 \alpha \beta \\ \\ & = {( \alpha + \beta )}^{2} - 4 \alpha \beta \\ \\ & = {(7)}^{2} - 4(3) \\ \\ & = 49 - 12 \\ \\ & = \quad \boxed{ \bf37} & \end{aligned}$

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$\mid \underline{\underline{\LARGE\bf\green{Brainliest \: Answer}}}\mid$

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