If $\frac{4}{27} = 2^{a} \cdot 3^{b}$ , then find the quadratic polynomial with zeroes 'a' and 'b'.

Open in App

Solution

Given,

$\frac{4}{27} = 2^{a} 3^{b}$

$\Rightarrow 2^{2} \cdot 3^{- 3} = 2^{a} 3^{b}$

$2^{2} = 2^{a} \Rightarrow a = 2$

$3^{- 3} = 3^{b} \Rightarrow b = - 3$

Therefore the required equation is,

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

$\Rightarrow x^{2} + x - 6 = 0$

Suggest Corrections

Similar questions

If a and b are the zeroes of the quadratic polynomial f(x)=3x^2-7x-6, find a polynomial whose zeroes are (i) a^2 and b^2 (ii) 2a+3b and 3a+2b

If ' $a$ ' and ' $b$ ' are the non-zero distinct zeros of $x^{2} + a x + b$ , then the least value of quadratic polynomial $x^{2} + a x + b$ is

(k - 1) x^{2} + k x + 1

2 a^{2} - 2 \sqrt{2} a + 1

4

5

Join BYJU'S Learning Program