Question

please help me everyone please

1.) x2 – 81 = 0
2.) 3t2 = 12
3.) (y − 4)2- 81 = 0

1.) Mrs Poe asked her carpenter to construct a square table with
an area of 16ft2. Can you tell the dimensions of the table?
Sheryl says that the solutions of the quadratic equations
w2 = 49 and w2 + 49 = 0 are the same. Do you agree with
Sheryl? Justify your answer.

Answer 1

1.

${x}^{2} - 81 = 0 \\ {x}^{2} = 81 \\ x = \sqrt{81} \\ x = 9$

2.

$3 {t}^{2} = 12 \\ {t}^{2} = \frac{12}{3} \\ {t}^{2} = 4 \\ t = \sqrt{4} \\ t = 2$

3.

$(y - 4)2 - 81 = 0 \\ 2y - 8 - 81 = 0 \\ 2y - 89 = 0 \\ 2y = 89 \\ y = \frac{89}{2} \\ y = 44.5$

1)

$Area = {(side)}^{2} \\ {(16ft)}^{2}={(side)}^{2} \\ 4ft = side$

A square has equal sides. So, each of the side of the square table is 4ft.

2)

${w}^{2} = 49 \\ {w} = \sqrt{49} \\ w = + 7 \: \: or \: \: - 7 \\ \\ {w}^{2} + 49 = 0 \\ {(w)}^{2} + {(7)}^{2} = 0 \\ (w + 7)(w - 7) = 0 \\ w + 7 = 0 \: \: or \: \: w - 7 = 0 \\ w = - 7 \: \: \: or \: \: \: w = + 7$

Therefore, Sheryl is right