Exam practice

$x^2 - 4x + 5 = 2x - 3$, so $x^2 - 6x + 8 = 0$. Yes: the highest power of x is 2.

$x^2 + x + 8 = x^2 - 4$, so x + 12 = 0. No: the $x^2$ terms cancel, leaving a linear equation.

$2x^2 + 3x = x^2 + 1$, so $x^2 + 3x - 1 = 0$. Yes: highest power 2.

$x^3 + 6x^2 + 12x + 8 = x^3 - 4$, so $6x^2 + 12x + 12 = 0$. Yes: the $x^3$ terms cancel and the highest power left is 2.

$x^2 + 3x + 1 = x^2 - 4x + 4$, so 7x − 3 = 0. No: linear.

$x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x$, so $x^3 - 6x^2 - 14x - 8 = 0$. No: the highest power is 3 (a cubic).

Product −10, sum −3: −5 and 2. (x − 5)(x + 2) = 0: x = 5 or −2.

ac = −12, sum 1: 4 and −3.
$2x^2 + 4x - 3x - 6 = 2x(x + 2) - 3(x + 2) = (x + 2)(2x - 3) = 0$: x = −2 or 3/2.

Multiply by 8: $16x^2 - 8x + 1 = 0$. Product 16, sum −8: −4 and −4.
$16x^2 - 4x - 4x + 1 = 4x(4x - 1) - 1(4x - 1) = (4x - 1)^2 = 0$: x = 1/4 (both roots equal).

Product 100, sum −20: −10 and −10.
$10x(10x - 1) - 1(10x - 1) = (10x - 1)^2 = 0$: x = 1/10 (both roots equal).

Product 324, sum −45: −36 and −9. (x − 36)(x − 9) = 0: x = 9 or 36.

Product 182, sum −27: −14 and −13. (x − 14)(x − 13) = 0: x = 13 or 14.

$x^2 - 6x + 3^2 = 0$, so $(x - 3)^2 = 0$ and x = 3 (both roots equal).

Divide by 9: $x^2 - \tfrac53 x = -\tfrac23$. Add $\left(\tfrac56\right)^2 = \tfrac{25}{36}$:
$$\begin{aligned} \left(x - \tfrac56\right)^2 &= \tfrac{25}{36} - \tfrac{24}{36} = \tfrac{1}{36} \\ x - \tfrac56 &= \pm \tfrac16 \end{aligned}$$
x = 1 or 2/3.

Divide by 2: $x^2 - \tfrac52 x = -\tfrac32$. Add $\left(\tfrac54\right)^2 = \tfrac{25}{16}$:
$$\begin{aligned} \left(x - \tfrac54\right)^2 &= \tfrac{25}{16} - \tfrac{24}{16} = \tfrac{1}{16} \\ x - \tfrac54 &= \pm \tfrac14 \end{aligned}$$
x = 3/2 or 1.

Divide by 5: $x^2 - \tfrac65 x = \tfrac25$. Add $\left(\tfrac35\right)^2 = \tfrac{9}{25}$:
$$\begin{aligned} \left(x - \tfrac35\right)^2 &= \tfrac{10}{25} + \tfrac{9}{25} = \tfrac{19}{25} \\ x - \tfrac35 &= \pm \tfrac{\sqrt{19}}{5} \end{aligned}$$
$x = \dfrac{3 + \sqrt{19}}{5}$ or $\dfrac{3 - \sqrt{19}}{5}$ (about 1.47 and −0.27).

$x^2 - 2x = -\tfrac{15}{16}$. Add $1^2$:
$$\begin{aligned} (x - 1)^2 &= 1 - \tfrac{15}{16} = \tfrac{1}{16} \\ x - 1 &= \pm \tfrac14 \end{aligned}$$
x = 5/4 or 3/4.

Add $\left(\tfrac13\right)^2 = \tfrac19$:
$$\begin{aligned} \left(x + \tfrac13\right)^2 &= \tfrac{35}{9} + \tfrac19 = \tfrac{36}{9} = 4 \\ x + \tfrac13 &= \pm 2 \end{aligned}$$
x = 5/3 or −7/3.

a = 1, b = −9, c = 20: $b^2 - 4ac = 81 - 80 = 1$.
$x = \dfrac{9 \pm 1}{2}$: x = 5 or 4.

a = 1, b = 2, c = −143: $b^2 - 4ac = 4 + 572 = 576 = 24^2$.
$x = \dfrac{-2 \pm 24}{2}$: x = 11 or −13.

a = 3, b = −5, c = 2: $b^2 - 4ac = 25 - 24 = 1$.
$x = \dfrac{5 \pm 1}{6}$: x = 1 or 2/3.

a = 2, $b = -2\sqrt2$, c = 1: $b^2 - 4ac = 8 - 8 = 0$.
$x = \dfrac{2\sqrt2 \pm 0}{4} = \dfrac{\sqrt2}{2}$: $x = \dfrac{1}{\sqrt2}$ (both roots equal; $\frac{1}{\sqrt2} = \frac{\sqrt2}{2} \approx 0.71$).

Multiply by x: $x^2 + 1 = 3x$, so $x^2 - 3x + 1 = 0$; $b^2 - 4ac = 9 - 4 = 5$.
$x = \dfrac{3 + \sqrt5}{2}$ or $\dfrac{3 - \sqrt5}{2}$ (about 2.62 and 0.38).

Multiply by x(x − 2): (x − 2) + x = 3x(x − 2), so $2x - 2 = 3x^2 - 6x$ and $3x^2 - 8x + 2 = 0$.
$b^2 - 4ac = 64 - 24 = 40$, $\sqrt{40} = 2\sqrt{10}$: $x = \dfrac{8 \pm 2\sqrt{10}}{6}$.
$x = \dfrac{4 + \sqrt{10}}{3}$ or $\dfrac{4 - \sqrt{10}}{3}$ (about 2.39 and 0.28).

The left side is $\dfrac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \dfrac{-11}{(x + 4)(x - 7)}$, so (x + 4)(x − 7) = −30.
$x^2 - 3x - 28 + 30 = 0$, so $x^2 - 3x + 2 = 0$; $b^2 - 4ac = 9 - 8 = 1$.
$x = \dfrac{3 \pm 1}{2}$: x = 2 or 1.

Let his Mathematics mark be x; his English mark is 30 − x.
$$\begin{aligned} (x + 2)(27 - x) &= 210 \\ 25x + 54 - x^2 &= 210 \\ x^2 - 25x + 156 &= 0 \\ (x - 12)(x - 13) &= 0 \end{aligned}$$
Mathematics 12 and English 18, or Mathematics 13 and English 17. Both fit: 14 × 15 = 210 and 15 × 14 = 210.

a. Let the shorter side be x m. By Pythagoras’ theorem:
$$\begin{aligned} x^2 + (x + 30)^2 &= (x + 60)^2 \\ 2x^2 + 60x + 900 &= x^2 + 120x + 3600 \\ x^2 - 60x - 2700 &= 0 \\ (x - 90)(x + 30) &= 0 \end{aligned}$$
x = −30 is rejected. Breadth 90 m, length 120 m (and the diagonal 150 m).

• b.Area 120 × 90 = 10800 m²; one turf is 36 m². 10800 ÷ 36 = 300 turfs (12 m along the length and 3 m across fit exactly: 10 × 30).
• c.Perimeter 2(120 + 90) = 420 m; 4 rounds are 1,680 m. 1680 × 15 = Rs 25,200.

$V = (x - 2)(x + 3)(x + 1) = x^3 + 2x^2 - 5x - 6$ cubic units (a cubic, not a quadratic: three lengths are multiplied).
For example, with x = 5: 3 × 8 × 6 = 144 cubic units.
Every side must be positive, and x − 2 is the smallest, so x must be more than 2.

The area (x + 5)(x − 3) is 0 when x = 3 or x = −5. At x = 3 the width is 0: the rectangle is squashed flat into a line 8 long. At x = −5 the length is 0 but the width is −8, which no rectangle can have, so only x = 3 makes sense. A product is zero exactly when one factor is zero: that is the factorization method in a picture.

$x^2 + 4x + 4 = (x + 2)^2 = 0$, so x = −2 (both roots equal). The square of side x + 2 then has side 0 and area 0. For any other x the area $(x + 2)^2$ is positive, which is why this equation has only one root.

The answer depends on your family. If you are 15, the grandmother is 4 × 15 = 60; if your mother is 40, the difference is 20 years. In general, with your age a and your mother’s m, the difference is 4a − m years (or m − 4a if your mother is older).

$x^2 + 11 = 36$, $x^2 = 25$, x = ±5. A natural number is positive: 5.

$x^2 - 11 = 25$, $x^2 = 36$: x = 6 or −6 (both work).

$2x^2 - 7 = 91$, $x^2 = 49$, x = ±7. The number is positive: 7.

$x^2 = 9$, x = ±3: 3 (natural).

$x^2 = 100$: x = 10 or −10.

$2x^2 - 17 = 55$, $2x^2 = 72$, $x^2 = 36$: x = 6 or −6.

$2x^2 = 288$, $x^2 = 144$, x = ±12: 12 (positive).

$x^2 + x - 72 = 0$, (x + 9)(x − 8) = 0. The number is positive: 8 (8 + 64 = 72 ✓).

x(x + 2) = 80, $x^2 + 2x - 80 = 0$, (x + 10)(x − 8) = 0.
8 and 10, or −10 and −8 (both pairs multiply to 80).

$x^2(x + 2)^2 = \{x(x + 2)\}^2$, and 225 is $15^2$.

$\{x(x + 2)\}^2 = 225$, so x(x + 2) = 15 or x(x + 2) = −15.
$x^2 + 2x - 15 = 0$ gives (x + 5)(x − 3) = 0, so x = 3 or −5. $x^2 + 2x + 15 = 0$ has no real root ($b^2 - 4ac = 4 - 60 < 0$).
3 and 5, or −5 and −3. Check: 9 × 25 = 225 ✓.

$x + \dfrac1x = \dfrac{10}{3}$, so $3x^2 - 10x + 3 = 0$, (3x − 1)(x − 3) = 0: x = 3 or 1/3.

$x^2 + (21 - x)^2 = 261$, so $2x^2 - 42x + 180 = 0$, $x^2 - 21x + 90 = 0$, (x − 6)(x − 15) = 0.
6 and 15 (36 + 225 = 261 ✓).

Younger x, elder x + 4: $x^2 + 4x - 221 = 0$, (x + 17)(x − 13) = 0, so x = 13.
17 years and 13 years.

x(22 − x) = 120, $x^2 - 22x + 120 = 0$, (x − 10)(x − 12) = 0.
12 years and 10 years.

x(x + 3) = 180, $x^2 + 3x - 180 = 0$, (x + 15)(x − 12) = 0, so x = 12.
15 years and 12 years.

(40 − x)(13 − x) = 198, $x^2 - 53x + 322 = 0$, (x − 7)(x − 46) = 0.
46 years ago the son wasn’t born, so 7 years ago (33 × 6 = 198 ✓).

(34 + x)(4 + x) = 400, $x^2 + 38x - 264 = 0$, (x + 44)(x − 6) = 0.
In 6 years (40 × 10 = 400 ✓).

(35 + x)(1 + x) = 240, $x^2 + 36x - 205 = 0$, (x + 41)(x − 5) = 0.
In 5 years (40 × 6 = 240 ✓).

(35 − x)(27 − x) = 425, $x^2 - 62x + 520 = 0$, (x − 10)(x − 52) = 0.
52 years ago neither was born, so 10 years ago (25 × 17 = 425 ✓).

$x^2 + (x + 17)^2 = 625$, $2x^2 + 34x - 336 = 0$, $x^2 + 17x - 168 = 0$, (x + 24)(x − 7) = 0.
7 m and 24 m (49 + 576 = 625 ✓).

Shortest side x, hypotenuse 2x + 6, third side 2x + 4:
$$\begin{aligned} x^2 + (2x + 4)^2 &= (2x + 6)^2 \\ 5x^2 + 16x + 16 &= 4x^2 + 24x + 36 \\ x^2 - 8x - 20 &= 0 \\ (x - 10)(x + 2) &= 0 \end{aligned}$$
10 m, 24 m and 26 m ($100 + 576 = 676 = 26^2$ ✓).

x + y = 25, xy = 150: $x^2 - 25x + 150 = 0$, (x − 15)(x − 10) = 0.
Length 15 m, breadth 10 m.

x + y = 15, xy = 54: $x^2 - 15x + 54 = 0$, (x − 9)(x − 6) = 0.
Length 9 m, breadth 6 m.

$24^2 + x^2 = (x + 16)^2$, so $576 + x^2 = x^2 + 32x + 256$. The $x^2$ terms cancel: 32x = 320, x = 10.
Area 24 × 10 = 240 m² (the diagonal is 26 m).

x + y = 90, xy = 2000: $x^2 - 90x + 2000 = 0$, (x − 50)(x − 40) = 0: length 50 m, breadth 40 m.
A square needs equal sides, so the length must come down to the breadth, 40 m: a cut of 10 m, and $\dfrac{10}{50} \times 100\% =$ 20%.

Tens digit x, units digit y: 10x + y = 4(x + y), so 6x = 3y and y = 2x.
10x + 2x = 3x · 2x, so $6x^2 - 12x = 0$, 6x(x − 2) = 0. The tens digit can’t be 0, so x = 2, y = 4.
24. Check: 4 × 6 = 24 ✓ and 3 × 8 = 24 ✓.

• a.$\dfrac{180}{x - 5} - \dfrac{180}{x} = 3$. Dividing by 3 and multiplying by x(x − 5): 60x − 60(x − 5) = x(x − 5), so $x^2 - 5x - 300 = 0$, (x − 20)(x + 15) = 0. 20 students.
• b.15 were present: 180 ÷ 15 = 12 pencils each (3 more than the planned 9).