find zeros if f (x) =x³-5x²-2x+24 and product is 12
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=> in the question
f(x) = x³ - 5x² - 2x + 24
And we have given that the product of it's two zeroes is 12 .
So, let the zeroes of the given cubic polynomial be φ , β and γ .
From the given condition we have,
φβ = 12 ..................(1)
and also we have ,
φ + β + γ = coefficient of x²/coefficient of x³ = 5 ......................(2)
φβγ = - constant term/ coefficient of x³ = -24 ...........................(3)
Substituting the value of φβ in equation no. (3) we get,
φβγ = -24
12γ = -24
γ = -24/12
γ = -2 ......................(4)
Putting the value of γ = -2 in equation no. (2) we get
φ + β + γ = 5
φ + β + (-2) = 5
φ + β = 5 +2
φ + β = 7 ...................(5)
Now, squaring on both sides we will get ,
(φ + β)² = (7)²
We know the expression of quadratic polynomial [ (φ + β)² = (φ - β)²+ 4φβ) ]
∴ ( φ - β )² + 4*12 = 49 [∵ φβ = 12 ]
(φ - β)² + 48 = 49
( φ - β)² = 49 - 48
(φ - β)² = 1
∴ φ - β = 1 .................(6)
Now, we will add the equation no. (5) and equation no. (6) we get,
φ + β = 7
φ - β = 1
----------------
2φ = 8
φ = 8/2
φ = 4
Putting φ = 4 in equation no. (5) we get,
φ - β = 1
4 - β = 1
- β = 1 - 4
-β = - 3 .......(multiplying by -1 on both side)
β = 3
∴ The zeroes are φ, β, γ = 4, 3, -2
this is your answer....
f(x) = x³ - 5x² - 2x + 24
And we have given that the product of it's two zeroes is 12 .
So, let the zeroes of the given cubic polynomial be φ , β and γ .
From the given condition we have,
φβ = 12 ..................(1)
and also we have ,
φ + β + γ = coefficient of x²/coefficient of x³ = 5 ......................(2)
φβγ = - constant term/ coefficient of x³ = -24 ...........................(3)
Substituting the value of φβ in equation no. (3) we get,
φβγ = -24
12γ = -24
γ = -24/12
γ = -2 ......................(4)
Putting the value of γ = -2 in equation no. (2) we get
φ + β + γ = 5
φ + β + (-2) = 5
φ + β = 5 +2
φ + β = 7 ...................(5)
Now, squaring on both sides we will get ,
(φ + β)² = (7)²
We know the expression of quadratic polynomial [ (φ + β)² = (φ - β)²+ 4φβ) ]
∴ ( φ - β )² + 4*12 = 49 [∵ φβ = 12 ]
(φ - β)² + 48 = 49
( φ - β)² = 49 - 48
(φ - β)² = 1
∴ φ - β = 1 .................(6)
Now, we will add the equation no. (5) and equation no. (6) we get,
φ + β = 7
φ - β = 1
----------------
2φ = 8
φ = 8/2
φ = 4
Putting φ = 4 in equation no. (5) we get,
φ - β = 1
4 - β = 1
- β = 1 - 4
-β = - 3 .......(multiplying by -1 on both side)
β = 3
∴ The zeroes are φ, β, γ = 4, 3, -2
this is your answer....
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Solution:-
=> The polynomial
x³-5x²-2x+24
=> and compare with
ax³+bx²+cx+d
=> α+β+y= -b/a =5
=> αβy = -d/a = -24
12y = -24....(as given the product
of two zeros..
i.e. αβ=12
y= -2.
=> α+β+y=5
=> α+β-2=5
=> α+β =7. .................(1)
=> (α+β)²=7²
=> (α-β)²+4αβ=49
=> (α-β)²+4*12=49
=> (α-β)²+48= 49
=> (α-β)² =1
=> α-β = √1
=> α-β=1. ....................(2)
• subtracting (2) From (1)
=> α+β-(α-β)=7-1
=> α+β-α+β=6
=> 2β =6
=> β=3
• putting the value of β in the (2) eq
=> α-3=1
=> α=4
=> α=4,β=3,y= -2
i hope it helps you.
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