Question

find the zeros of the polynomial and verify the relation between zeros and coefficients t2-16

Answer 1

Answer:

he graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

     Class_10_Maths_Polynomials_Division_Of_Graphs          

Answer:

(i) The graph does not intersect the x axis, so there is no zero in this graph.

(ii) The graph intersects the x axis at one place, so there is only one zero in this graph.

 

(iii) The graph intersects the x axis at three places, so there are three zeros in this graph.

(iv) The graph intersects the x axis at two places, so there are two zeros in this graph.

(v) The graph intersects the x axis at four places, so there are four zeros in this graph.

(vi) The graph intersects the x axis at three places, so there are three zeros in this graph.

 

                                                                   Exercise 2.1

Question 1:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x2 – 2x – 8                                (ii) 4s2 – 4s + 1                               (iii) 6x2 – 3 – 7x

(iv) 4u2 + 8u                                 (v) t2 – 15                                        (vi) 3x2 – x – 4

Answer:

(i) x2 – 2x – 8                        

= x2 – 4x + 2x - 8

= x(x - 4) + 2(x - 4)

= (x - 4)(x + 2)

The value of x2 – 2x – 8 is zero if x + 2 = 0 or x – 4 = 0

So, x = -2, 4

Therefore zeroes of x2 – 2x – 8 are -2 and 4

Now, Sum of zeroes = -2 + 4 = 2 = -(-2)/1 = -(Coefficient of x)/ (Coefficient of x2)  

Product of zeroes = (-2) * 4 = -8 = -8/1 = Constant term/ (Coefficient of x2)      

(ii) 4s2 – 4s + 1                              

= 4s2 – 2s – 2s + 1

= 2s(2s - 1) - 1(2s - 1)

= (2s - 1)(2s - 1)

The value of 4s2 – 4s + 1 is zero if 2s - 1 = 0

So, s = 1/2, 1/2

Therefore zeroes of 4s2 – 4s + 1 are 1/2 and 1/2

Now, Sum of zeroes = 1/2 + 1/2 = 1 = -(4)/4 = -(Coefficient of x)/ (Coefficient of x2)  

Product of zeroes = 1/2 * 1/2 = 1/4 = Constant term/ (Coefficient of x2)

(iii) 6x2 – 3 – 7x

= 6x2 – 7x - 3

= 6x2 – 9x + 2x - 3

= 3x(2x - 3) + 1(2x - 3)

= (2x - 3)(3x + 1)

The value of 6x2 – 7x – 2 is zero if 2x - 3 = 0 or 3x + 1 = 0

So, x = -1/3, 3/2

Therefore zeroes of 6x2 – 7x – 3 are -1/3 and 3/2

Sum of zeroes = -1/3 + 3/2 = (-2 + 9)/6 = 7/6 = -(-7)/6 = -(Coefficient of x)/ (Coefficient of x2)  

Product of zeroes = (-1/3) * 3/2 = -1/2 = -3/6 = Constant term/ (Coefficient of x2)

(iv) 4u2 + 8u

= 4u(u + 2)

The value of 4u2 + 8u is zero if 4u = 0 or u + 2 = 0

=> u = 0, -2

Therefore, the zeroes of 4u2 + 8u are 0 and -2.

Now, Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of x)/ (Coefficient of x2)  

Product of zeroes = 0 * (-2) = 0 = 0/4 = Constant term/ (Coefficient of x2)

(v) t2 – 15  

= t2 – (√15)2                      

= (t - √15)(t + √15)

The value of t2 – 15 is zero if t - √15 or t + √15 = 0

=> t = √15, -√15

Therefore, the zeroes of t2 – 15 are √15 and -√15.

Now, Sum of zeroes = -√15 + √15 = 0 = -(0)/1 = -(Coefficient of x)/ (Coefficient of x2)  

Product of zeroes = (-√15) * √15 = -15 = -15/1 = Constant term/ (Coefficient of x2)

(vi) 3x2 – x – 4

= 3x2 – 4x + 3x - 4

= x(3x - 4) + 1(3x - 4)

= (3x - 4)(x + 1)

The value of 6x2 – 7x – 2 is zero if 3x - 4 = 0 or x + 1 = 0

So, x = 4/3, -1

Therefore zeroes of 3x2 – x – 4 are 4/3 and -1

Sum of zeroes = 4/3 + (-1) = (4 - 3)/3 = 1/3 = -(-1)/3 = -(Coefficient of x)/ (Coefficient of x2)  

Product of zeroes = 4/3 * 1 = -4/3 = Constant term/ (Coefficient of x2)

Question 2:

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.  

(i) 1/4, -1              (ii) √2, 1/3             (iii) 0, √5            (iv) 1, 1         (v) -1/4, 1/4            (vi) 4, 1

Answer:

(i) Let α and β are the zeroes of the quadratic polynomial ax2 + bx + c.

Given, α + β = 1/4 = -b/a

and α * β = -1 = -4/4 = c/a

On comparing, we get

a = 4, b = -1, c = -4

Hence, the required quadratic polynomial is 4x2 - x - 4.

(ii) Let α and β are the zeroes of the quadratic polynomial ax2 + bx + c.

Given, α + β = √2= 3√2/3 = -b/a

and α * β = 1/3 = c/a

On comparing, we get

a = 3, b = -3√2, c = 1

Hence, the required quadratic polynomial is 3x2 - 3√2x + 1.

(iii) Let α and β are the zeroes of the quadratic polynomial ax2 + bx + c.

Given, α + β = 0 = 0/1 = -b/a

and α * β = √5 = √5/1 = c/a

On comparing, we get

a = 1, b = 0, c = √5

Hence, the required quadratic polynomial is x2 – 0*x + √5 = x2 + √5.

                   

 

 

     

 

 

 

Answer 2

Answer With Step-by-step explanation:

$t^{2} - 16\\= t^{2} - 4^{2} \\= (t-4)(t+4)\\zeros of the polynomial are 4, -4\\\\VERIFICATION:\\\alpha +\beta =4+(-4)=0 =\frac{-b}{a} \\\alpha \beta =4(-4)=(-16) =\frac{c}{a} \\\\\\hence proved\\\\\\\\please mark me brainliest$