Question

if one zero of the polynomial 2x²-5x-(2k+1) is twice the other. find both the zeros of the polynomial and the value of k​

Answer 1

Answer:

Let the zeroes be a and 2a.

Sum of zeroes,

$a \: + 2a = \frac{5}{2} \\ 3a = \frac{5}{2} \\ a = \frac{5}{6}$

Product of the roots,

$a \times 2a = \frac{ - 2k - 1}{2} \\ 2 {a}^{2} = \frac{ - 2k - 1}{2} \\ 2 \times \frac{5}{6} \times \frac{5}{6} = \frac{ - 2k - 1}{2} \\ \frac{25}{9} = - 2k - 1 \\ 2k = \frac{ - 34}{9} \\ k = \frac{ - 17}{9}$

Answer 2

Answer :

zeros are 5/6 and 5/3

k = -17/9

Note:

★ The possible values of the variable for which the polynomial becomes zero are called its zeros .

★ A quadratic polynomial can have atmost two zeros .

★ The general form of a quadratic polynomial is given as ; ax² + bx + c .

★ If α and ß are the zeros of the quadratic polynomial ax² + bx + c , then ;

• Sum of zeros , (α + ß) = -b/a

• Product of zeros , (αß) = c/a

★ If α and ß are the zeros of a quadratic polynomial , then that quadratic polynomial is given as : k•[ x² - (α + ß)x + αß ] , k ≠ 0 .

Solution :

Here ,

The given quadratic polynomial is ;

2x² - 5x - (2k + 1)

Now ,

Comparing the given quadratic polynomial with the general quadratic polynomial ax² + bx + c , we have ;

a = 2

b = -5

c = -(2k + 1)

Also ,

It is given that , one of the zero of the given polynomial is twice the other .

Thus ,

Let z and 2z be the two zeros of the given quadratic polynomial .

Now ,

The sum of zeros will be ;

=> z + 2z = -b/a

=> 3z = -(-5)/2

=> z = 5/(2•3)

=> z = 5/6

Also ,

2z = 2•(-5/6) = 5/3

Hence ,

The zeros of the given quadratic polynomial are 5/6 and 5/3 .

Also ,

The product of zeros will be ;

=> z•2z = c/a

=> 2z² = -(2k + 1)/2

=> 2•(5/6)² = -(2k + 1)/2

=> 2•25/36 = -(2k + 1)/2

=> 25/9= -(2k + 1)

=> 25 = -9(2k + 1)

=> 25 = -18k - 9

=> 18k = -9 - 25

=> 18k = -34

=> k = -34/18

=> k = -17/9

Hence k = -17/9 .