Question

form the quadratic polynomials whose zeroes are 1/3 and 1/4​

Answer 1

Hello Samya

Here is your answer below,,

Zeros are given 1/3 and 1/4

Now going to solution,

$x = \frac{1}{3} \\ = > x - \frac{1 }{3} \\ or \\ x = \frac{1}{4} \\ = > x - \frac{1}{4} \\ \\ so \: that \: \\ \\ (x - \frac{1}{3} )(x - \frac{1}{4} ) = 0\\ = > x {}^{2} - \frac{1}{4} x - \frac{1}{3} x + \frac{1}{12} = 0 \\ = > \frac{12x {}^{2} - 3x - 4x + 1}{12} = 0 \\ = > 12x {}^{2} - 7x + 1 = 0 \times 12\\ = > 12x {}^{2} - 7x + 1 = 0$

I hope that helps you,

Please follow me and Mark as Brainlist

Answer 2

Answer:

hello samiya

Step-by-step explanation:

the correct answer is 12x^2 + 5x - 3