Comparing the given polynomial ax² + bx + (2a)/(b) with the standard quadratic form A x² + B x + C, we identify the coefficient of x² as A = a, the coefficient of x as B = b, and the constant term as C = (2a)/(b). For a valid quadratic polynomial, a ≠ 0 and b ≠ 0.

Comparing the given polynomial ax² + bx + (2a)/(b) with the standard quadratic form A x² + B x + C, we identify the coefficient of x² as A = a, the coefficient of x as B = b, and the constant term as C = (2a)/(b). For a valid quadratic polynomial, a ≠ 0 and b ≠ 0.

Let the two zeroes of the quadratic polynomial be α and β. Since the zeroes are reciprocals of each other, we have β = (1)/(α), which gives their product as α · β = α · (1)/(α) = 1.

If the zeroes of polynomial $ax^2+bx+2a/b$ . Are reciprocal of each other ,then the value of b is

Answer: $b = 2$

Step-by-step solution

Step 1: Identify coefficients of the quadratic polynomial

Comparing the given polynomial $ax^2 + bx + \frac{2a}{b}$ with the standard quadratic form $A x^2 + B x + C$ , we identify the coefficient of $x^2$ as $A = a$ , the coefficient of $x$ as $B = b$ , and the constant term as $C = \frac{2a}{b}$ . For a valid quadratic polynomial, $a \neq 0$ and $b \neq 0$ .

Step 2: Use the reciprocal property of the zeroes

Let the two zeroes of the quadratic polynomial be $\alpha$ and $\beta$ . Since the zeroes are reciprocals of each other, we have $\beta = \frac{1}{\alpha}$ , which gives their product as $\alpha \cdot \beta = \alpha \cdot \frac{1}{\alpha} = 1$ .

Step 3: Equate the product of zeroes to find b

For any quadratic polynomial, the product of the zeroes is given by the ratio of the constant term to the leading coefficient, $\frac{C}{A}$ . Substituting $C = \frac{2a}{b}$ and $A = a$ , we get $\frac{2a/b}{a} = \frac{2}{b} = 1$ . Solving for $b$ yields $b = 2$ .

More from this chapter

All Polynomials solutions →

Solve your own maths question free →

If the zeroes of polynomial ax2+bx+2a/bax^2+bx+2a/bax2+bx+2a/b. Are reciprocal of each other ,then the value of b is