Inequalities

Rearrange so one side is 0, factorise, then sketch the parabola and decide whether you want where it is above or below the x-axis.

1. Make one side zero: x^2+3x-28\geqslant 0
2. Factorise: (x+7)(x-4)\geqslant 0
3. Critical values: x=-7 and x=4
4. y=x^2+3x-28 is a U-shaped curve crossing the x-axis at -7 and 4.
5. \geqslant 0 means on or above the x-axis: the two outer parts.
6. Answer: x\leqslant -7 or x\geqslant 4
7. (-7\leqslant x\leqslant 4 is where the curve is below the axis; x\geqslant 4 misses the second region; x\leqslant -4 or x\geqslant 7 comes from factorising with the wrong signs.)

Expand both brackets carefully (watch -2\times -1), then collect the x terms on the side where they stay positive.

1. Expand: 7-6x+2<4x+24
2. Simplify the left side: 9-6x<4x+24
3. Add 6x to both sides: 9<10x+24
4. Subtract 24: -15<10x
5. Divide by 10: -1.5<x
6. Answer: x>-\dfrac{3}{2}
7. (If you divide by a negative number instead, remember to reverse the inequality sign.)

Area of a parallelogram is base times perpendicular height. Once you have solved the quadratic inequality, check which values of x make every length positive.

Part (a)

1. Area of a parallelogram = base \times perpendicular height
2. (x+4)(3x-2)<24
3. Expand: 3x^2-2x+12x-8<24
4. 3x^2+10x-8<24
5. Subtract 24: 3x^2+10x-32<0

Part (b)

1. Factorise: (3x+16)(x-2)<0
2. Critical values: x=-\dfrac{16}{3} and x=2
3. The curve is U-shaped, so it is below the x-axis between the roots: -\dfrac{16}{3}<x<2
4. But every length must be positive.
5. Height: 3x-2>0, so x>\dfrac{2}{3} (base: x+4>0 gives x>-4, which is weaker)
6. Combine: \dfrac{2}{3}<x<2