Question

1. If one zero of a quadratic polynomial \( \left(k x^{2}+3 x+k\right) \) is 2 , then the what is the value of \( k \) ? 

2. The graph of a polynomial is shown in Figure. What is the number of its zeroes? 

3. Find the quadratic polynomial, the sum of whose zeroes is \( -5 \) and their product is 6.

4. If one zero of the polynomial \( \left(3 x^{2}+8 x+k\right) \) is the reciprocal of the other, then what is the value of \( k \) ? 

5. What is the value of \( x \), for which the polynomials \( x^{2}-1 \) and \( x^{2}-2 x+1 \) vanish simultaneously? 

6. Find a quadratic polynomial, whose zeroes are \( -3 \) and 4 ? 

7. If the sum of the zeroes of the polynomial \( p(x)=2 x^{3}-3 k x^{2}+4 x-5 \) is 6 , then what is the value of \( k \). 

8. In given figure, the graph of a polynomial \( p(x) \) is shown. Calculate the number of zeroes of \( p(x) \). 

9. If zeroes of the polynomial \( x^{2}+4 x+2 a \) are \( a \) and \( \frac{2}{a} \) then find the value of \( a \). 

10. If one of the zeroes of the quadratic polynomial \( p(x)=14 x^{2}-42 k^{2} x-9 \) is negative of the other, find the value of ' \( k \) '. 

11. If \( \alpha \) and \( \beta \) are zeroes of the polynomial \( p(x)=x^{2}-x-k \), such that \( \alpha-\beta=9 \), find \( k \). 

12. If \( \alpha \) and \( \beta \) are zeroes of \( x^{2}-(k-6) x+2(2 k-1) \), find the value of \( k \) if \( \alpha+\beta=\frac{1}{2} \alpha \beta \). 

13. Quadratic polynomial \( 2 x^{2}-3 x+1 \) has zeroes as a and \( b \). Now form a quadratic polynomial whose zeroes are \( 3 a \) and \( 3 b \). 

14. Find the zeroes of the quadratic polynomial \( 5 x^{2}+8 x-4 \) and verify the relationship between the zeroes and the coefficients of the polynomial.

15. Find the value for \( k \) for which \( x^{4}+10 x^{3}+25 x^{2}+15 x+k \) is exactly divisible by \( x+7 \). 

16. If \( a \) and \( b \) are the zeroes of polynomial \( p(x)=3 x^{2}+2 x+1 \), find the polynomial whose zeroes are \( \frac{1-a}{1+a} \) and \( \frac{1-b}{1+b} \). 

17. If \( \beta \) and \( \frac{1}{\beta} \) are zeroes of the polynomial \( \left(a^{2}+a\right) x^{2}+61 x+6 a \). Find the value of \( \beta \) and

Answer 1

1. If one zero of a quadratic polynomial (kx² + 3x + k) is 2, then the value of k can be found by substituting x = 2 into the polynomial and setting it equal to zero:

k(2)² + 3(2) + k = 0

4k + 6 + k = 0

5k + 6 = 0

5k = -6

k = -6/5

So the value of k is -6/5.

3. The quadratic polynomial with sum of zeroes equal to -5 and product of zeroes equal to 6 can be obtained by using Vieta's formulas. Let the zeroes be denoted as a and b. The sum of the zeroes is given by a + b = -5, and the product of the zeroes is ab = 6.

To find the quadratic polynomial, we need to form an equation with the given information:

(x - a)(x - b) = 0

Expanding this equation:

x² - (a + b)x + ab = 0

Substituting the values -5 for (a + b) and 6 for ab:

x² + 5x + 6 = 0

Therefore, the quadratic polynomial is x² + 5x + 6.

4. If one zero of the polynomial (3x² + 8x + k) is the reciprocal of the other, then the value of k can be determined by considering the relationship between the zeroes. Let the zeroes be denoted as a and 1/a.

From the given information, we have:

a * (1/a) = 1

This implies that a² = 1.

Since we know that a² = 1, the polynomial can be factored as follows:

(3x² + 8x + k) = 3(x - a)(x - 1/a)

Expanding this equation:

3(x² - (a + 1/a)x + k/a) = 0

Comparing the coefficients, we can equate them to the corresponding coefficients of the original polynomial:

-3(a + 1/a) = 8 (coefficient of x)

k/a = 0 (constant term)

From the first equation, we have:

-3(a^2 + 1) = 8

-3(1 + 1) = 8

-3(2) = 8

-6 = 8 (Contradiction)

Since the equation leads to a contradiction, there is no valid value of k that satisfies the given condition.

5. To find the value of x for which the polynomials x² - 1 and x² - 2x + 1 vanish simultaneously, we need to set both polynomials equal to zero:

x² - 1 = 0

x² - 2x + 1 = 0

Simplifying each equation:

x² = 1

x² - 2x + 1 = 0

The first equation, x² = 1, has two solutions: x = 1 and x = -1.

The second equation, x² - 2x + 1 = 0, is a perfect square and can be factored as (x - 1)² = 0. This equation has a repeated root, x = 1.