Pure Mathematics Practice: Equations and Inequalities — Answers

Questions: Equations and Inequalities Drills.

Question 1

(a) The critical values are $x=-5$ and $x=2$. The product is negative between the roots, so

$$-5<x<2.$$

(b) The critical values are $x=-2$ and $x=4$. The value $x=-2$ is excluded because the expression is undefined there; $x=4$ is included because it makes the expression zero. A sign chart gives

$$x<-2\text{or}x\ge 4.$$

(c) An expression containing $x$ may be positive on some intervals, negative on others and zero at excluded values. Multiplication by a negative quantity reverses an inequality, while multiplication by zero can destroy equivalence. For example, $x+2$ changes sign at $x=-2$ and is zero where the original rational expression is undefined.

Question 2

(a)

$$2x-3=5\text{or}2x-3=-5,$$

so

$$x=4\text{or}x=-1.$$

(b)

$$|x+1|<4⟺-4<x+1<4,$$

so

$$-5<x<3.$$

(c) Both sides are non-negative, so squaring preserves the inequality:

$$(x-2{)}^2\ge (2x+1{)}^2.$$

Hence

$$3x^2+8x-3\le 0⟺(3x-1)(x+3)\le 0.$$

Therefore

$$-3\le x\le \frac{1}{3}.$$

Question 3

From the first equation,

$$y=2x+z-5.$$

Substitution into the second and third equations gives

$$5x+z=11,5x+3z=17.$$

Subtracting yields $2z=6$, so $z=3$. Then

$$x=\frac{8}{5},y=2\left(\frac{8}{5}\right)+3-5=\frac{6}{5}.$$

For example, checking in $x+2y-z=1$:

$$\frac{8}{5}+2\left(\frac{6}{5}\right)-3=1.$$

Thus

$$(x,y,z)=\left(\frac{8}{5},\frac{6}{5},3\right).$$

Question 4

(a)

$$\Delta =(-2k{)}^2-4(1)(9)=4(k^2-9).$$

(b) The coefficient of $x^2$ is positive. Therefore $q(x)>0$ for every real $x$ precisely when the quadratic has no real root:

$$\Delta <0⟺k^2<9.$$

Hence

$$-3<k<3.$$

The strict inequality is important: at $k=\pm 3$, the quadratic touches the $x$-axis and is zero at one real value.

(c) For $k=4$,

$$q(x)=x^2-8x+9,$$

whose roots are

$$x=4\pm \sqrt{7}.$$

Since the parabola opens upwards,

$$4-\sqrt{7}\le x\le 4+\sqrt{7}.$$