Question

The zeroes of the quadratic polynomial 3x^2-48 are

Answer 1

Answer:

$let \: \: \: \: \: \: {3x}^{2} - 48 = 0 \\ \\ divide \: both \: side \: by \: 3 \: then \: we \: get \: \\ \\ {x}^{2} - 16 = 0 \\ \\ on \: comparing \: the \: given \: equation \: with \: {ax}^{2} + bx + c = 0 \: \: we \: get \: \\ a = 1 \: \: \: \: \: b = 0 \: \: \: \: c = - 16 \\ \\ wkt \: \\ {b}^{2} - 4ac \: = {(0)}^{2} - 4 \times 1 \times - 16 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = + 64 \\ {b}^{2} - 4ac \: > 0 \\ \\ x = \frac{ - b \frac{ + }{ - } \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ x = \frac{ - 0 \frac{ + }{ - } \sqrt{64} }{2 \times 1} \\ \\ x = \frac{ \frac{ + }{ - }8 }{2} \\ \\ \times = \frac{ + }{ - } 4$

x = 4 and x = - 4 are the zeroes of the given quadratic equation .......

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Answer 2

Answer:

Step-by-step explanation:

3x^2=48

divide both side by 3

3x^2=/3=48/3

x^2=16

x=root 16

x=±4

thats it!!!!

hope my answer was helpful :)