Class 10 Math All topics Polynomials

If $α$ and $β$ are the zeros of the polynomial $f(x)=6x_{2}−3−7x$, then $(α+1)(β+1)$ is equal to

(a)

$25 $

(b)

$35 $

(c)

$52 $

(d)

$53 $

Correct answer: (b)

Solution: If $α$, $β$ are the zeros of the quadratic polynomial $f(x)=6x_{2}−3−7x$, then

$(α+β )=a−b =−6(−7) =67 ,αβ =ac =6−3 =2−1 $

Now, $(α+1)(β+1)$

$=αβ+α+β+1$

$=2−1 +67 +1$

$=6−3+7+6 $

$=610 $

$=35 $

We know that equation $ax_{2}+bx+c=0$

Then sum of roots $=a−b $ and product of roots$=ac $

$(α+β )=a−b =−6(−7) =67 ,αβ =ac =6−3 =2−1 $

Now, $(α+1)(β+1)$

$=αβ+α+β+1$

$=2−1 +67 +1$

$=6−3+7+6 $

$=610 $

$=35 $

Similar topics

introduction to trigonometry

functions

some applications of trigonometry

quadratic equations

polynomials

introduction to trigonometry

functions

some applications of trigonometry

quadratic equations

polynomials

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