Question

if alfa and bitta are the zeroes of the polynomial f(x)=6x²-3-7x then (alfa+1) (bitta+1) is equal to​

Answer 1

Given :-

we have a polynomial

• 6x² - 7x - 3 = 0

To Find :-

• value of (alpha + 1) (beta + 1)

Solution :-

firstly compare the polynomial with ax² + bx + c = 0

and we know that,

$\boxed {\bold{sum \: of \: zeroes = \frac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} } }}$

and

$\boxed{\bold{ product \: of \: zeroes = \frac{constant \: term}{coefficient \: of \: {x }^{2} } }}$

and here, in 6x² - 7x - 3 = 0

• coefficient of x² = 6
• coefficient of x = -7
• constant term = -3

zeroes of the polynomial are α and β

$\bold{: \implies \alpha + \beta = \frac{ - ( - 7)}{6} = \frac{7}{6} } .......(i)$

and

$\bold{: \implies \alpha \beta = \frac{ - 3}{6} = - 3}.......(ii)$

now, we have to find value of

$( \alpha + 1)( \beta + 1) \\ \\ \bold{: \implies \alpha \beta + \alpha + \beta + 1 }$

put following values from eq.(i) and (ii)

• α + β = 7/6
• αβ = -3

$\bold{: \implies \frac{7}{6} - 3 + 1 } \: \: \\ \\ \bold{: \implies \frac{4 - 18 + 6}{6} } \\ \\ \bold{: \implies \cancel \frac{8}{6} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bold{: \implies \frac{4}{3} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

so, value of (α + 1)(β + 1) is 4/3