Question

Find all the zeroes of 2x⁴ - 3x³ - 3x² + 6x - 2, if two of its zeroes are √2 and -√2. ​

Answer 1

Answer:

q(x)=2x4-3x3-3x2+6x-2 

q(x)=p(x)⋅(x-√2)(x+√2) 

q(x)=p(x)⋅(x2-2) 

bu using identity (x+a)(x-a)=x2-a2 

p(x)=q(x)x2-2 

=2x4-3x3-3x2+6x-2x2-2 

by dividing , we get p(x)=2x2-x+1 

=2x2-2x-x+1 

=2x2-2x-x+1 

=2x[x-1]-1[x-1] 

p(x)=(x-1)(2x-1) 

x=1,12 

answer

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Answer 2

Answer:

√2 , -√2 , ½ , 1

Step-by-step explanation:

given that,

the quadratic equation,

2x⁴ - 3x³ - 3x² + 6x - 2

given zeros = √2 and -√2

so,

roots of the quadratic equation

x = √2

x - √2 = 0

also,

x = -√2

x + √2 = 0

now,

we have roots of the quadratic equation

(x - √2) and (x + √2)

now,

multiply the both roots,

(x - √2) (x + √2)

x² - 2

so,

by multiplying both roots we get,

x² - 2

now,

divide the quadratic equation by the multiplication of the roots,

by dividing the quadratic equation by the multiplication of roots,

we get the quotient

2x² - 3x + 1

split the middle term,

2x² - 3x + 1

2x² - 2x - x + 1

2x(x - 1) -1(x - 1)

(2x - 1)(x - 1) = 0

2x - 1 = 0

2x = 1

x = 1/2

also,

x - 1 = 0

x = 1

so,

all the zeroes of the quadratic equation

= √2 , -√2 , ½ , 1