Please can you help? (inequalities as level)

[liverpool204]

Hi im struggling a lot with this maths, i have no idea what it's asking haha, you've got to choose one of the forms and then work out what p and q are for each but how do you find that? Thanks in advance.

1) The solution to the inequality

x2−5x+4>0
is written in the form

x < p or x > q
or
p < x < q
with p = and q =

2) The solution to the inequality
2x^2−x−3>0
is written in the form

x < p or x > q
or
p < x < q
with p = and q =

3)The solution to the inequality

8+2x−x^2<0
is written in the form

x < p or x > q
p < x < q

with p =
and q =

4)The solution to the inequality

6x^2+x−2≥0
is written in the form

x ≤ p or x ≥ q
p ≤ x ≤ q
with p = and q=

5)The solution to the inequality

2x^2−x+1>x^2−4x−1
is written in the form

x < p or x > q
p < x < q
with p = and q =

6)The quadratic equation

x2+(k−1)x+4=0
has at least one real root.

The range of possible values of k is written in the form

x < p or x > q
p < x < q
x ≤ p or x ≥ q
p ≤ x ≤ q
with p = and q =


Thanks in advance, i just have no idea on what p and q are in inequalities and quadratics.

 

[HallsofIvy]

These are all quadratic inequalities and I imagine that you know that a quadratic equation can have two solutions. In general, a quadratic, \(\displaystyle ax^2+ bx+ c\) can be written as \(\displaystyle a(x- x_0)(x- x_1)\) where \(\displaystyle x_1\) and \(\displaystyle x_1\) are the solutions to the equation \(\displaystyle ax^2+ bx+ c= 0\).

So a quadratic inequality, like \(\displaystyle ax^2+ bx+ c> 0\) can be written as \(\displaystyle a(x- x_0)(x- x_1)> 0\). Now, a product of two numbers is either positive or negative depending on the signs of the numbers. The product of two positive or two negative numbers is positive while the product of two numbers of different signs is negative. Of course, whether we want the product \(\displaystyle (x- x_0)(x- x_1)\) positive or negative depends on the sign of a but that is easy to see. And the sign of \(\displaystyle (x_x0)(x- x_1)\) is easy- write the two numbers, \(\displaystyle x_0\) and \(\displaystyle x_1\) in order so that \(\displaystyle x_0< x_1\). If x< x_0, then both \(\displaystyle x- x_0\) and \(\displaystyle x- x_1\) are "smaller number minus larger number" and so both \(\displaystyle x- x_0\) and \(\displaystyle x- x_1\) are negative and their product is positive. If \(\displaystyle x_0< x< x_1\), that is, if x is between \(\displaystyle x_0\) and \(\displaystyle x_1\) then \(\displaystyle x- x_0\) is positive while \(\displaystyle x- x_1\) is still negative- their product is negative. Finally, if \(\displaystyle x> x_1\) both \(\displaystyle x- x_0\) and \(\displaystyle x- x_1\) are positive so their product is positive again.

For example, if you are asked to solve \(\displaystyle x^2- 5x+ 6> 0\). First solve \(\displaystyle x^2- 5x+ 6= 0\). That's easy, \(\displaystyle x^2- 5x+ 6\) factors: \(\displaystyle x^2- 5x+ 6= (x- 3)(x- 2)\). That is, \(\displaystyle x^2- 5x+ 6= 0\) when x= 2 or x= 3. If x< 2, then both x- 2 and x- 3 are negative so \(\displaystyle x^2- 5x+ 6\) is positive. If 2< x< 3, x- 2 is positive while x- 3 is still negative so \(\displaystyle x^2- 5x+ 6\) is negative. Finally, if x> 3, both x- 2 and x- 3 are positive so \(\displaystyle x^2- 5x+ 6\) is positive again.

Even if the quadratic doesn't factor easily, you can use the quadratic formula or complete the square to find the roots of the quadratic equation and get the factors that way.