Question

Hey genius ji here is ur question?

IF 2 ZEROES OF THE POLYNOMIAL
$xto \: the \: power4 - 6xcube - 26x sq. + 138x - 35$
are
$2 + or - \sqrt3$
find other zeroes.

PLS HELP ME GENIUS JI ​

Answer 1

x⁴ - 6x³ - 26x² + 138x - 35 = 0

Now

ã(alpha)+ ß(beta) + ¥ (gamma) + d(delta) = - b / a

à + ß + ¥ + d = -(-6) / 1

à + ß + 2 - √3 = 6

à + ß = 6 - 2 + √3

à + ß = 4 + √3 ...(1)

Now

àߥd {product of zeroes) = e / a

àß2×(-√3) = - 35 / 1

àß × (-2√3) = - 35

àß = - 35 / -2√3

àß = 35 / 2√3

à = 35 / 2√3 × ß...(2)

put the value of à in equation (1)

35/2√3 + ß = 4 + √3

ß = 4 + √3 - 35 / 2√3

ß = 8√3 - 6 - 35 / 2√3

ß = 8√3 - 29 / 2√3

ß = 4 - 29/2√3

put the value of ß in equation (2)

à = 35/2√3 × 4 - 29 /2√3

à = (35 × 4) - (35 × 29) / 12

à = 140 - 1015 / 12

à = 875 / 12

Hence the zeroes are

4 - 29 / 2√3 and 875 / 12

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Answer 2

Answer:

Step-by-step explanation:

Given :

Two zeroes of the polynomial $x^4 - 6x^3 - 26x^2 + 138x - 35 = 0$ are 2 ±√3.

To Find :

The other zeroes of the polynomial,.

Solution :

The sum of the zeroes of the polynomial of the form,

$ax^4 + bx^3 + cx^2 + dx + e = 0$ is $\frac{-b}{a}$

here,  a = 1 , b = -6 , c = -26 , d = 138 , e = -35

Let the other 2 zeroes be x and y,

Sum of the zeroes,

(2 + √3) + (2 - √3) + x + y = $\frac{-(-6)}{1}$

4 + x + y = 6

x + y = 2,   ...(i)

_

The product of the zeroes is $\frac{e}{a}$

(2 + √3)(2 - √3)(x)(y) = $\frac{-35}{1}$

(2² - (√3)²)xy = -35

( 4 - 3) xy = -35 ⇒ xy = -35 ..(ii)

_

( x + y ) ² = x² + y² + 2xy

⇒ (2)² = x² + y² + 2( -35 )

⇒ 4 = x² + y² - 70

⇒ x² + y² = 74  ...(iii)

_

(x - y)² = x² + y² - 2xy

⇒ ( x - y )² = 74 - 2( -35 )

⇒ ( x - y )² = 74 + 70 = 144

⇒ ( x - y ) = √144 = 12

⇒ x - y = 12 ...(iv)

Adding, (i) & (iv)

We get,

(x + y) + (x - y) = 2 + 12 = 14

⇒ 2x = 14 ⇒ x = 7,.

_

Substituting the value of x in (i),

We get,

7 + y = 2

⇒ y = -5

∴ The two zeroes are 7,-5