Question

if α and β are the zeroes of the Quadratic polynomial p(x)= 6x²+x-2, Find the value of 1/α + 1/β - αβ​

Answer 1

Answer:5/6

Step-by-step explanation:

Let’s first solve the equation to get our zeroes of the equation i.e. alpha and beta.

6x2+x−2=0

6x2+4x−3x−2=0

2x(3x+2)−(3x+2)=0

which gives us two equations,

2x−1=0=>x=1/2(alpha)

3x+2=0=>x=−2/3(beta)

So, adding the values in question

• 1/(alpha) + 1/(beta) - ( alpha )( beta )
• => -3/2 + 2/1 - ( -2/3 x 1/2 )
• => -3+4/2 - ( -1/3 )
• => 1/2 + 1/3

= 5/6

Answer 2

Answer:

$\boxed{\sf \dfrac{1}{\alpha}+\dfrac{1}{\beta}-\alpha\beta =\dfrac{-5}{6}}$

Step-by-step explanation:

Given that , α and β are the Zeroes of the quadratic polynomial 6x² + x + 2 . And we need to find the value of 1/α + 1/β - αβ.

Now with respect to Standard form of a quadratic polynomial ax² + bx + c , we have a = 6 , b = 1 and

c = 2 .

With respect to Standard form , the sum of zeroes and product of zeroes is given by ,

$\boxed{\begin{array}{c}\sf Sum\ of\ Zeroes =\red{\dfrac{-b}{a} } \\\\ \sf Product \ of\ Zeroes=\red{\dfrac{c}{a}}\end{array}}$

• Hence here :-

• Sum = -1/6
• Product = 2/6 = 1/3 .

• Now we need to simplify ,

$\sf\dashrightarrow \dfrac{1}{\alpha}+\dfrac{1}{\beta}-\alpha\beta \\\\\sf\dashrightarrow \dfrac{ \alpha + \beta }{\alpha\beta}-\alpha\beta \\\\\sf\dashrightarrow \dfrac{\dfrac{-1}{6}}{\dfrac{1}{3}}-\dfrac{1}{3} \\\\\sf\dashrightarrow \dfrac{-1}{6}\times 3 - \dfrac{1}{3} \\\\\sf\dashrightarrow \dfrac{-1}{2}-\dfrac{1}{3} \\\\\sf\dashrightarrow \dfrac{ -3-2}{6} \\\\\sf\dashrightarrow \boxed{\pink{\sf \dfrac{-5}{6}}}$

$\rule{200}2$