Quadratic Inequalities

The two factors must have opposite signs for their product to be negative.

1. The product is zero at x=2 and x=5.
2. Between these values, x-2 is positive and x-5 is negative.

3. Therefore the solution is 2<x<5; the strict inequality excludes both roots.

Find the roots and include them if the inequality allows equality.

1. The roots are -3 and 1.
2. The product is negative between them and zero at the roots.

3. Hence -3\le x\le1.

Think about negative values of x too: (-5)^2 is bigger than 16, but (-3)^2 is not.

1. The boundary equation x^2=16 has roots -4 and 4.
2. Squares are less than 16 between these values.

3. The solution is -4<x<4.

A positive quadratic is positive outside its two roots.

1. Factorise: x^2-9=(x-3)(x+3). The roots are -3 and 3.
2. Outside the roots, both factors have the same sign, so the product is positive.

3. The solution is x<-3 or x>3, excluding the roots.

Factorise first, then decide where the quadratic is negative or zero.

1. Factorise: x^2+x-6=(x+3)(x-2), so the critical values (where it equals 0) are x=-3 and x=2.
2. Sketch y=x^2+x-6: it is a U-shaped curve crossing the x-axis at -3 and 2.

3. x^2+x-6\le0 means the curve is on or below the x-axis, which happens between the critical values.
4. The sign is \le, so the critical values are included: -3\le x\le2.

Factorise and check whether the required values lie between or outside the roots.

1. The quadratic factorises as (x-2)(x-5).
2. It is \ge0 outside the roots 2 and 5, including the roots themselves.

3. Therefore x\le2 or x\ge5.

Find a factorisation with one factor starting with twice the unknown.

1. Factorise: 2x^2-5x-3=(2x+1)(x-3).
2. The roots are -\frac12 and 3.
3. The quadratic is negative between its roots, so -\frac12<x<3.

Rearrange so that the squared term is positive before finding the boundary values.

1. Adding x^2 and subtracting 3 gives x^2<9.
2. The boundary values are -3 and 3.

3. The solution is -3<x<3.

You can multiply by negative one, but must reverse the inequality sign.

1. Multiply by -1 to get x^2-4x-5\le0.
2. Factorise: (x-5)(x+1)\le0.
3. The quadratic is \le0 between the roots, and at the roots themselves, giving -1\le x\le5.

A square cannot be negative, so consider when it is zero.

1. For real x, the square (x-2)^2 is always at least zero.
2. To be at most zero, it must equal zero: x-2=0.
3. The only solution is x=2.

Solve the quadratic inequality over the real numbers before listing its integers.

1. Factorise to get (x-4)(x+2)<0.
2. The real solutions lie strictly between the roots: -2<x<4.

3. The integer solutions are \{-1,0,1,2,3\}.

The roots may be fractions even though the coefficients are integers.

1. Factorise: 3x^2+8x+4=(3x+2)(x+2).
2. The roots are -\frac23 and -2; in increasing order they are -2,-\frac23.
3. The quadratic is \le0 between and at the roots, so -2\le x\le-\frac23.

Move every term to one side before factorising.

1. Subtract 12 to get x^2+4x-12<0.
2. Factorise: (x+6)(x-2)<0.
3. The product is negative between its roots, so -6<x<2.

The quantity being squared must lie between the negative and positive square roots of the bound.

1. The inequality is equivalent to -3\le x-4\le3.
2. Add 4 to each part.
3. The solution is 1\le x\le7.

The curve is below the line where x^2-4x-12 is less than 9.

1. Below the line means x^2-4x-12<9.
2. Rearrange and factorise: x^2-4x-21=(x-7)(x+3).
3. This expression is negative between the roots, giving -3<x<7.

Solve each quadratic inequality on its own, then find the values that are in both.

1. Solve each inequality separately, then find the values of x that satisfy both.
2. First: x^2-3x-10=(x-5)(x+2), with critical values -2 and 5. The graph is U-shaped, so it is on or below the axis between them: -2\le x\le5.

3. Second: 2x^2+x-6=(2x-3)(x+2), with critical values -2 and \frac32. This U-shaped graph is above the axis outside them: x<-2 or x>\frac32.

4. x<-2 has no values in -2\le x\le5, and x=-2 fails the second inequality (2x^2+x-6=0 there, which is not more than 0).
5. The values in both are \frac32<x\le5.

Expand both sides and collect all terms on one side before factorising.

1. Expanding gives 2x^2-5x-3>x^2-2x+1.
2. Collect terms to get x^2-3x-4>0.
3. Factorise: (x-4)(x+1)>0.
4. The product is positive outside its roots, so x<-1 or x>4.

Form a quadratic inequality from the area, then use the integer and length restrictions.

1. The area condition is (n+2)(n-1)\le54.
2. Expand and rearrange to obtain n^2+n-56\le0.
3. Factorise as (n+8)(n-7)\le0, giving -8\le n\le7.

4. With n>1 and integer, the greatest possible value is 7. The area at this value is 9\times6=54\text{ cm}^2.

Solve each inequality, then list the positive integers that satisfy both.

1. Linear inequality: 4n-3<25 gives 4n<28, so n<7.
2. Quadratic inequality: rearrange to n^2-2n-8>0 and factorise: (n-4)(n+2)>0. The critical values are -2 and 4.
3. The graph of y=n^2-2n-8 is U-shaped, so it is above the axis outside the critical values: n<-2 or n>4.

4. n is positive, so we need 4<n<7. The integers are 5 and 6.

5. Check n=4: 16-8=8, which is not more than 8, so 4 is not allowed.

Find the values allowed by each bound on the square, then take their overlap.

1. The upper bound x^2\le25 gives -5\le x\le5.
2. The strict lower bound x^2>4 requires x<-2 or x>2.
3. Keep the overlap of these conditions.

4. The result is -5\le x<-2 or 2<x\le5.