find the solution sets of the ff. quadratic inequalities if possible.?

Question

1. x squared >49
2. 3y squared - 5 >0
3. x squared - 6x < 0
4. 5t squared - 12 > 0
5. x squared > -32

Answer 1

1. x>7 or x<-7
2. y>sqrt(5/3) or y< -sqrt(5/3)
3. 0 4. t>sqrt(12/5) or t< -sqrt(12/5)
5. All Real numbers

Let's look at one of them so you know how to solve it.
Number 4.
5t^2 - 12 > 0

5t^2 > 12 (add 12)

t^2 > 12/5 (divide by 5. be careful when */ by a negative.)

t>sqrt(12/5) or t< -sqrt(12/5)
(square root of both sides. remember the negative solution. a negative farther from zero, when squared, gives a larger value!)

Answer 2

x^2 > 49 =>x >7 or x < -7
(3y)^2-5 >0 => (3y)^2 >5 =>y > (5^1/2)/3 or < -5(^1/2)/3
x^2-6x <0 =>x(x-6) <0
=>x>0 and x <6
(5t)^2 -12 >0 => t >(12^1/2)/5 or <-(12^1/2)/5
x^2 >-32 =>x^2+32 >0 which is true for all x

Answer 3

You managed to make 2 blunders that canceled out! x (4-x) ? 4 = x (4-x) - 4 ? 0 (incorrect) - (xx - 4x + 4) ? 0 = xx - 4x + 4 ? 0 (incorrect) contained in the first case you shouldn't have replaced the inequality signal, contained in the 2d case you need to have. So contained in the authentic you get: (x-2) (x-2) ? 0 imagine about what y=(x-2)(x-2) appears like. that's a favorable (upwards dealing with) parabola. It has a double root (x=2) so that you know it touches the x axis at this aspect. for this reason you know that each aspect on that parabola might want to be >=0. for this reason the answer set is all actual numbers! inspect the plot contained in the link lower than to get a seen illustration. The shaded area (i.e the comprehensive determination line) is the answer set.

Answer 4

1. the range x>7 & the range x<-7
2. the range y>sqrt(5/3) & the range y < -sqrt(5/3)
3. x(x-6)<0 ==> x<0& x-6 >0 Or x>0 & x-6<0 the first answer is refused then the answe is 6>x>0
4. the range t>sqrt(2.4) & the range t<-sqrt(2.4)
5. all real numbers getsthe equation

Answer 5

1.x^2>9
x>7 or <-7

2.3y^2-5>0
3y^2>5
y^2>5/3
y>(5/3)^1/2 or <(-5/3)

3.x^2-6x<0
x(x-6)<0
x>0 and x<6

4.5t^2-12>0
5t^2>12
t^2>12/5
t>(12/5)^1/2 or t<(-12/5)

5.x^2>-32
x element of real nos

Answer 6

x<-7 and x>7
y<-root5/3 and y>root5/3
x<0 and x<6
t<-root12/5 and t>root12/5
x<-4 root2 i and x>4 root2 i