Math, asked by reshmachauhan2908, 8 months ago

Find the zeroes of 12+28 x-5x²

Answers

Answered by ronaldoChristiano

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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
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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

Thanks !!!! ✌✌✌✌☺

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Answered by Anonymous

Question:-

Find the zeroes of

$12 + 28 - 5 {x}^{2}$

Answer:-

$x = 6 \: or \: x = \frac{ - 2}{5}$

Solution:-

$12 + 28x - 5 {x}^{2} = 0\\ - (5 {x}^{2} - 28x - 12) = 0 \\ - (5 {x}^{2} - 30x + 2x - 12) = 0 \\ - [5x(x - 6) + 2(x - 6) ]= 0 \\ - (x - 6)(5x + 2) = 0 \\ (x - 6)(5x + 2) = 0 \\ x = 6 \: or \: x = \frac{ - 2}{5}$

Hope its help uh❤

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