Question

If α and β are the zeroes of the polynomial f(x) = 6x² - 3 - 7x, then (α + 1) (β + 1) is equal to?

Answer 1

$\huge\bf{Question:-}$

If α and β are the zeroes of the polynomial f(x) = 6x² - 3 - 7x, then (α + 1) (β + 1) is equal to?

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$\huge\bf{Solution:-}$

6x^2 - 7x - 3 = 0

ax^2 + bx + c = 0

here;

a = 6

b = -7 and

c = -3

alfha + beta = -b/a = -(-7)/6 = 7/6

alfha × beta = c/a = -3/6 = -1/3

x^2 + (alpha + beta)x + alpha × beta

2 alpha + 2 beta = 2(alpha + beta) = 2×7/6 =7/3

2 alpha × 2 beta = 4 alpha × beta = 4×-1/2 = -2

x^2 -7/3x -2 = 0

x^2 +(-7x-6)/3 = 0

x^2 -7x -6 = 0×3

x^2 -7x -6 = 0

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Answer 2

Given :- α and β are the zeroes of the polynomial f(x) = 6x² - 3 - 7x

To Find :- The value of "( α + 1 ) ( β + 1 )".

Used Concepts :- A general Quadratic polynomial is in the form of "ax² + bx + c". Where ,

Sum of zeroes of the polynomial is given by :- -b/a .

Product of the zeroes of the polynomial is given by:-

c/a.

Solution :-

The given polynomial is :-

6x² - 7x -3

Here , a = 6 , b = - 7 and c = - 3

So , Sum of Zeroes = α + β = -b/a = - ( - 7 ) / 6 = 7/6 .

Product of Zeroes = αβ = c/a = -3/6 = - 1/2

Now ,

( α + 1 ) ( β + 1 )

α ( β + 1 ) + 1 ( β + 1 )

αβ + α + β + 1

αβ + ( α + β ) + 1

Now ,Putting all values we get,

$\frac{ - 1}{2} + ( \frac{7}{6} ) + 1$

$\frac{ - 3 + 7 + 6}{6}$

$\frac{13 - 3}{6}$

$\frac{10}{6}$

$\frac{5}{3}$

Therefore, The required answer is 5/3.