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

1 thought on “Find the zeros of the following<br />1) 6x²-7x-3<br />2)4x²-4+1​”

1. Answer:

   Required Answer :–

   [tex] {6x}^{2} – 7x – 3[/tex]

   [tex] = {6x}^{2} + 2x – 9x – 3[/tex]

   [tex] = 2x(3x + 1) – 3(3x + 1)[/tex]

   [tex] = (2x – 3)(3x + 1)[/tex]

   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 :–

   [tex]2x – 3 = 0[/tex]

   [tex]2x = 3[/tex]

   [tex]x = \frac{3}{2} [/tex]

   Now, let’s find second zero :–

   [tex]3x + 1 = 0[/tex]

   [tex]3 x = – 1[/tex]

   [tex]x = \frac{ – 1}{3} [/tex]

   The zeroes are 3/2 and -1/3

   Second quadratic polynomial solution

   2.4x²-4+1

   [tex]4x²-4x+1[/tex]

   [tex] = 4 {x}^{2} – 2x – 2x + 1[/tex]

   [tex] = 2x(2x – 1) – 1(2x – 1)[/tex]

   [tex] = (2x – 1)(2x – 1)[/tex]

   Finding zeroes :–

   [tex]2x – 1 = 0[/tex]

   [tex]2x = 1[/tex]

   [tex]x = \frac{1}{2} [/tex]

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

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