Question

BRAINLIEST UR ANS AWARDED DONT MAKE SPAM​

Answer 1

Given that a and ß are zeros of quadratic polynomial f(x) = x² + x - 2

We have to find value of (1/a - 1/ß)

Solution

Given quadratic polynomial :

→ f(x) = x² + x - 2

Now,

→ x² + 2x - x - 2 = 0

→ x(x + 2) - 1(x + 2) = 0

→ (x - 1)(x + 2) = 0

→ (x - 1) = 0 or, (x + 2) = 0

→ x = 1 or, x = -2

∴ Value of a = 1

∴ Value of ß = -2

Now going to the required equation :

→ 1/a - 1/ß

Putting values of a and ß :

→ 1/1 - 1/(-2)

→ 1 + 1/2

→ (2 + 1)/2

→ 3/2

∴ Required value of (1/a - 1/ß) = 3/2

Answer 2

$\red {\bf{\underline {Given:-}}}$

▪$\tt{\alpha \: and \: \beta \: are \: the \: zeros \: of \: the \: quadratic \: polynomial \:f(x) = { {x}^{2} + x - 2} }$

$\red {\bf{\underline {To\:Find:-}}}$

▪$\tt{The \: value \: of \: \frac{1}{ \alpha } - \frac{1}{ \beta } }$

$\huge\red {\bf{\underline {Solution:-}}}$

According to the question,

$\tt {{x}^{2} + x - 2 = 0}$

$⇒ \tt {{x}^{2} + 2x - x - 2 = 0}$

$⇒\tt {x(x + 2) - (x + 2) = 0}$

$⇒\tt {(x + 2)(x - 1) = 0}$

$⇒\tt {(x + 2) = 0 \: \: or \: \: (x - 1) = 0}$

$⇒\tt {x = - 2 \: \: or \: \: x = 1}$

$\tt {∴ \alpha = - 2 \: \: and \: \: \beta = 1}$

Now,

$\tt { \frac{1}{ \alpha } - \frac{1}{ \beta } }$

$⇒ \tt {\frac{1}{ - 2} - \frac{1}{1} }$

$⇒ \tt{\frac{1}{- 2} - 1}$

$⇒\tt { \frac{ 1 + 2}{-2}}$

$⇒ \tt {\frac{ 3}{-2}}$

$\bold\pink{\bf {Answer: -\frac{3}{2}}}$