Question

If alpha and beta are the zeroes of x^2+7x+12 then find the value of 1/alpha+1/beta-3alpha beta

step by step explanation needed to be marked as brainlist

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

If α and β are the zeroes of x²+7x+12 = 0 , then find the value of ( 1/α + 1/β - 3αβ ).

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{Given}}}}$

• Equation x²+7x+12 = 0
• α and β are two zeroes

$\Large{\underline{\mathfrak{\bf{Find}}}}$

• Value of ( 1/α + 1/β - 3αβ )

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

We know,

$\small\sf{\boxed{\boxed{\red{\:Sum\:of\:Zeroes\:=\:\dfrac{-(Coefficient\:of\:x)}{(Coefficient\:of\:x^2)}}}}}$

$:\mapsto\sf{\:Sum\:of\:Zeroes\:=\:\dfrac{-(7)}{1}}$

$:\mapsto\mathfrak{\bf{\:(\alpha+\beta)\:=\:-7}.....(1)}$

again,

$\small\sf{\boxed{\boxed{\red{\:Product\:of\:Zeroes\:=\:\dfrac{(constant\:part}{(Coefficient\:of\:x^2)}}}}}$

$:\mapsto\sf{\:product\:of\:Zeroes\:=\:\dfrac{12}{1}}$

$:\mapsto\mathfrak{\bf{\:(\alpha \beta)\:=\:12.......(2)}}$

Now,

$:\mapsto\sf{\orange{\:\dfrac{1}{\alpha}+\dfrac{1}{\beta}-3\alpha \beta}}$

$:\mapsto\sf{\:\dfrac{(\alpha+\beta)}{\alpha \beta}-3\alpha \beta}$

$\:\:\:\:\small\sf{\: keep\:value\:by\:equ(1)\:and\:(2)}$

$:\mapsto\sf{\:\dfrac{(-7)}{(12)}-3\times 12}$

$:\mapsto\sf{\:\dfrac{(-7-36\times 12)}{12}}$

$:\mapsto\sf{\:\dfrac{(-7-432)}{12}}$

$:\mapsto\mathfrak{\bf{\:\dfrac{-439}{12}\:\:\:\:\:Ans}}$

Answer 2

$\huge\bold\green{Answer:-}$

Given:

and beta are the zeroes of x^2+7x+12

Find:

we need to find the value of 1/alpha+1/beta-3alpha beta.

Solution:

Given polynomial is x² +7x +12

Comparing the given polynomial by:

ax² + bx + c

we get,

a =1

b =7

c =12

Sum of zeroes = -b/a

= -7/1

= -7

α + β = -7

Now, product of zeroes = c/a

= 12/1

= 12

αβ = 12

Now, inorder to find 1/α + 1/β -3αβ we have,

1/α + 1/β -3αβ

= (α + β)/αβ - 3αβ

= [(-7)/12] - 3(12)

= (-7)/12 - 36

= [(-7) -(432)]/12

= $\sf\dfrac{-439}{12}$