find the value of zeros?

Dear student,
For constant to be 20, it will not satisfy the given condition.
Assuming the eqn to be:
Given: x³ - 6x² + 3x + 10
Zeroes are of the form = a , a + b , a + 2b
Using relationship of coefficient and Zeroes of the polynomial we get
Sum of the roots =

\frac{- c o e f f i c i e n t o f x ²}{c o e f f i c i e n t o f x ³}

⇒ a + 2b + a + a + b = $\frac{- (- 6)}{1} = 6$
⇒ 3a + 3b = 6
⇒ 3 ( a + b ) = 6
⇒ a + b = 2 ..................(1)
Product of roots = $\frac{- c o n s \tan t}{c o e f f i c i e n t o f x ³}$

⇒ ( a + 2b )( a + b ) a = $\frac{- 10}{1} = - 10$

⇒ ( a + b + b )( a + b ) a = -10

⇒ ( 2 + b )( 2 ) a = -10 ( From equation (1) )

⇒ ( 2 + b ) 2a = -10

⇒ ( 2 + 2 - a ) 2a = -10

⇒ ( 4 - a ) 2a = -10

⇒ 4a - a² = -5

⇒ a² - 4a - 5 = 0

Now solving obtained quadratic equation we get,

a = 5 , -1

Now, Putting value of a in equation in Eqn (1)

We get,

When

a = 5 ⇒ 5 + b = 2 ⇒ b = -3

a = -1 ⇒ -1 + b = 2 ⇒ b = 3

when a = 5 and b = -3, Zeroes are 5 , ( 5 + (-3) ) = 2 , ( 5 + 2(-3) ) = -1

when a = -1 and b = 3 , Zeroes are -1 , ( -1 + 3 ) = 2 , ( -1 + 2(3) ) = 5

Therefore, Value of a = 5 & b = -3 or a = -1 & b = 3 and Zeroes are -1 , 2 , 5.
Regards

View Full Answer