Question

plz sent the ans plz ​

Answer 1

Step-by-step explanation:

9)

Given zero = 5

We know that the zero of ax+b is -b/a

So if the zero is 5 then the Polynomial is x-5

The required Polynomial is x-5

10)

Given zeroes are 1 and -1

Let α = 1 and β = -1

sum of the zeroes α + β = 1+(-1) = 1-1=0

Product of the zeroes αβ=1(-1) = -1

We know that

The Quadratic Polynomial whose zeroes α and β is K[x^2-(α + β )x +(αβ)]

=> K[x^2-(0)x+(-1)]

=> K[x^2-0x-1]

=> K[x^2-1]

If k =1 then the required Polynomial is x^2-1

11)

Given quardratic polynomial is

P(x) = 4x^2-5x-1

On Comparing this with the standard quardratic polynomial ax^2+bx+c

a = 4

b= -5

c=-1

We Know that

Sum of the zeroes α+ β = -b/a

α+ β = -(-5)/4

α+ β = 5/4

Product of the zeroes αβ = c/a

=> αβ = -1/4

Now,

α^2+ β^2

=>( α+ β )^2 -2( α β )

=> (5/4)^2-2(-1/4)

=> (25/16)+(1/2)

=> (25+8)/16

=> 33/16

The value of α^2+ β^2 = 33/16

12)

Given quardratic polynomial is x^2-x-6

To get zeroes we can write x^2-x-6= 0

=> x^2-3x+2x-6 = 0

=> x(x-3)+2(x-3) = 0

=> (x-3)(x+2) = 0

=> x-3 = 0 or x+2 = 0

=> x = 3 or x=-2

The zeroes are 3 and -2

One positive and another is negative