Question

Find the zeros of the following
1) 6x²-7x-3
2)4x²-4+1​

Answer 1

Answer:

Required Answer :-

${6x}^{2} - 7x - 3$

$= {6x}^{2} + 2x - 9x - 3$

$= 2x(3x + 1) - 3(3x + 1)$

$= (2x - 3)(3x + 1)$

So, the value of 6x²-7x-3 is zero when 2x-3=0 or 3x+ 1= 0,i.e when x= 3/2 or x = -1/3

Finding first zero :-

$2x - 3 = 0$

$2x = 3$

$x = \frac{3}{2}$

Now, let's find second zero :-

$3x + 1 = 0$

$3 x = - 1$

$x = \frac{ - 1}{3}$

The zeroes are 3/2 and -1/3

Second quadratic polynomial solution

2.4x²-4+1

$4x²-4x+1$

$= 4 {x}^{2} - 2x - 2x + 1$

$= 2x(2x - 1) - 1(2x - 1)$

$= (2x - 1)(2x - 1)$

Finding zeroes :-

$2x - 1 = 0$

$2x = 1$

$x = \frac{1}{2}$

so the two zeroes are 1/2 and 1/2.