Question

plzz answer fast i will mark as brainliest

Answer 1

1)
$35 {y}^{2} + 12 \div {y}^{2} = 44 \\ 35 {y}^{4} + 12 = 44 {y}^{2} \\ let \: {y}^{2} = x \\ 35 {x}^{2} - 44x + 12 = 0 \\ 35 {x}^{2} - 30x - 14x + 12 = 0 \\ 5x(7x - 6) - 2(7x - 6) = 0$
(7x-6) (5x-2) =0
7x-6=0 or 5x-2=0
x=6÷7 or x=2÷5
${y}^{2} = 0.85714 \: \: \: \:or \: \: \: \: {y }^{2} = 0.4$
y=0.92581 or y=0.63245

2)
${x}^{2} + 12 \div {x}^{2} = 7 \\ {x}^{4} + 12 = 7 {x}^{2} \\ let \: {x}^{2} = y \\ {y}^{2} - 7y + 12 = 0 \\ {y}^{2} - 3y - 4y + 12 = 0 \\ y(y - 3) - 4(y - 3) = 0$
(y-3) (y-4) =0
y-3=0 or y-4=0
y=3 or y= 4
${x}^{2} = 3 \: \: \: \: \: or \: \: \: \: {x}^{2} = 4$
x=1.732050 or x= 2

3)
$let \: {x}^{2} = y \: \\ 3 {y}^{2} - 13y + 10 = 0 \\ 3 {y}^{2} - 3y - 10y + 10 = 0 \\ 3y(y - 1) - 10(y - 1) = 0 \\ (y - 1)(3y - 10) = 0 \\ y = 1 \: \: \: \: or \: \: \: 3y = 10 \\ {x }^{2} = 1 \: \: \: \: \: or \: \: \: \: {x}^{2} = 3.333 \\ x = 1 \: \: \: \: or \: \: \: \: x = 1.82574$

4) [tex]2{y}^{2} + 15 \div {y}^{2} = 12 \\ 2{y}^{4} + 15 = 12 {y}^{2} \\ let \: x = {y}^{2} \\ 2 {x}^{2} - 12x + 15 = 0 \\ using \: formula \\ a = 2 \: \: \: \: \: \: \: b = - 12 \: \: \: \: \: \: c = 15 \\ x = - b (\: \: plus \: or \: minus) + - \sqrt{ {b}^{2} - 4ac \ } \div 2a \\ x = 12 + - \sqrt{ {12}^{2} - 4(2)(15) } \div 2(2) \\ x = 12 + - \sqrt{144 - 120} \div 4 \\ x = 12 + - \sqrt{24 } \div 4 \\ x = 12 + - 4.898979 \div 4 \\ x = 12 + 4.89 \div 4 \: \: \: or \: \: \: x = 12 - 4.89 \div 4 \\ x = 4.22 \: \: \: \: \: or \: \: \: \: \: x = 1.775 \\ y = 2.05 \: \: \: \: or \: \: \: \: y = 1.33