Question

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Answer 1

Answer:

p^2 - 4q.

Step-by-step explanation:

We know,

Quadratic equation which are written in the form of x^2 - Sx + P = 0 represent the sum & product of their roots as S and P.

So,

if a and b are roots then a + b = - p and ab = q.

Thus,

= > ( a - b )^2

= > a^2 + b^2 - 2ab

= > a^2 + b^2 + 2ab - 2ab - 2ab

= > ( a + b )^2 - 4ab

= > ( - p )^2 - 4( q )

= > p^2 - 4q.

Answer 2

Question

If $\alpha$ and $\beta$ are the roots of the equation x² + px + q = 0, then ($(\alpha - \beta)^2$ = ____

Solution

Comparing the given quadratic equation x² + px + q = 0 with standard equation ax² + bx + c = 0, we get

a = 1

b = p

c = q

Now, we know that sum of zeroes = -b/a and product of zeroes = c/a

This gives

$\alpha + \beta = \frac{-p}{1}=-p$

and $\sf\alpha\beta=\frac{q}{1} = q$

Now,

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

We know that x² + y² = (x + y)² - 2xy

Hence, $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$

So,

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

$\implies (\alpha - \beta)^2 = (\alpha + \beta)^2 - 2\alpha\beta-2\alpha\beta$

$\implies (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$

Put value of $(\alpha + \beta)$ and $\alpha\beta$

$(\alpha-\beta)^2 = (-p)^2 - 4(q)$

$(\alpha-\beta)^2 = p^2 - 4q$

Answer : p² - 4q