Question

If alpha,beeta are the zeroes of the polynomial,f(x)= x^2-x-4, then the value of alpha+beeta-alpha*beeta=?​

Answer 1

GIVEN :-

• ɑ and β are the zeroes of the polynomial f(x)= x² - x - 4.

TO FIND :-

• The value of (ɑ + β) - (ɑ × β).

SOLUTION :-

$: \implies \displaystyle \sf \: \alpha + \beta = \frac{ - \: co - efficient \: of \: x}{co - efficient \: of \: x ^{2} }$

$: \implies \displaystyle \sf \: \alpha + \beta = \frac{ - b}{a}$

$: \implies \displaystyle \sf \: \alpha + \beta = \frac{ -(-1)}{1}$

$: \implies \displaystyle \sf \: \boldsymbol{ \alpha + \beta } = 1$

Now,

$\\ \\ : \implies \displaystyle \sf \: \alpha \times \beta = \frac{constant \: term}{co - efficient \: of \: x ^{2} }$

$: \implies \displaystyle \sf \: \alpha \times \beta = \frac{c}{a}$

$: \implies \displaystyle \sf \: \alpha \times \beta = \frac{ - 4}{1}$

$: \implies \displaystyle \sf \: \boldsymbol{\alpha \times \beta} = - 4$

Now according to question we have to find the value of (ɑ + β) - (ɑ × β) . So,

$\\ \\ : \implies \displaystyle \sf \: ( \alpha + \beta ) - ( \alpha \times \beta ) = 1 - ( - 4)$

$: \implies \displaystyle \sf \: ( \alpha + \beta ) - ( \alpha \times \beta ) = 1 + 4$

$: \implies \underline{ \boxed{ \displaystyle \sf \: \boldsymbol{( \alpha + \beta ) - ( \alpha \times \beta ) = 5}}}$

Answer 2

Given:,

$\sf\qquad\bullet \alpha \: and \: \beta are \: zeroes \: of \: f(x) = {x}^{2} - x - 4$

Find:

$\sf\qquad\bullet (\alpha) + (\beta) - (\alpha) \times ( \beta )$

Solution:

Compare f(x) = x² - x - 4 = 0

with ax² + bx + c = 0

So, here

• a = 1
• b = -1
• c = -4

we, know that

$\boxed{ \sf Sum \: of \: Zeroes = \frac{ - b}{a} }$

where,

• a = 1
• b = -1

So,

$\hookrightarrow\sf Sum \: of \: Zeroes = \frac{ - b}{a}$

$\hookrightarrow\sf ( \alpha ) + ( \beta ) = \frac{ - ( - 1)}{1}$

$\hookrightarrow\sf ( \alpha ) + ( \beta ) = \frac{1}{1}$

$\hookrightarrow\sf ( \alpha ) + ( \beta ) = 1$

$\rule{287.5}{1.6} \\ \rule{287.5}{1.6}$

we, know that

$\boxed{ \sf Product \: of \: Zeroes = \frac{c}{a} }$

where,

• a = 1
• c = -4

So,

$\hookrightarrow\sf Product \: of \: Zeroes = \frac{c}{a}$

$\hookrightarrow\sf ( \alpha ) \times ( \beta ) = \frac{ - 4}{1}$

$\hookrightarrow\sf (\alpha) \times (\beta) = - 4$

So, we have to find

$\sf \dashrightarrow(\alpha) + (\beta) - (\alpha) \times ( \beta )$

where,

• $\alpha$ + $\beta$ = 1
• $\sf\alpha \times \beta = -4$

So,

$\sf \dashrightarrow(\alpha) + (\beta) - (\alpha) \times ( \beta )$

$\sf \dashrightarrow 1 - (-4)$

$\sf \dashrightarrow 1 + 4$

$\sf \dashrightarrow 5$

$\rule{287.5}{1.6} \\ \rule{287.5}{1.6}$

Hence, $\sf(\alpha) + (\beta) - (\alpha) \times ( \beta ) = 5$