[Admin]

in reply to: Rationalising The Denominator #10259

Rationalise $$\frac{2\sqrt2-3}{4+\sqrt{2}}$$.

$$=\frac{2\sqrt{2}-3}{4+\sqrt{2}}\times \frac{4-\sqrt{2}}{4-\sqrt{2}}$$

$$=\frac{8\sqrt2-4-12+2\sqrt2}{16-2}$$

$$=\frac{10\sqrt2-16}{14}$$

$$=\frac{5\sqrt2-8}{7}$$

[Admin]

in reply to: Trapezoidal Rule #10188

Use the Trapezoidal Rule with 4 sub-intervals to find an approximation for the area under the curve y = x⁴ + x between x = 0 and x = 4.

x          0     1     2     3     4
f(x)     0     2    18   84  260

h = 1 and the values of f(a) = 0 and f(b) =260

$$h=\frac{4-0}4=1$$

≈ ¹/₂{0 +240 + 2(2 + 18 + 84)}

≈ 234u²

[Admin]

in reply to: Equations #10113

Solve log₃ x = -2

$$3^{log_3x}=3^{-2}$$  

remember that $$a^{log_a x}$$ will simplify to x as the a to the power of log_a will cancel out

x = ¹/₉

 

[Admin]

in reply to: Log Laws #10071

If x = ln 2 and y = ln 3, find ln 6e in terms of x and y

ln 6e = ln 6 + ln e
= ln (2 × 3) + 1
= ln 2 + ln 3 + 1
= x + y + 1

[Admin]

in reply to: Challenging Equations #10010

Solve x^1/2– 6x^-1/2 – 5 = 0.

$$\sqrt x  – \frac{6}{{\sqrt x }} – 5 = 0$$

x – 6 – 5√x = 0

x –  5√x – 6 = 0

let y =  √x

y² – 5y – 6 = 0

(y – 6)(y + 1) = 0

y = 6, y = -1

√x = 6, √x = -1

x = 36, x = 1

[Admin]

in reply to: Maxima/Minima – Application Problems #9933

A piece of wire is 20cm long. It is cut into two portions, the length of one portion being x cm. Each portion is then bent to form a rectangle in which the length is twice the breadth.
a. Show that the sum of the areas of the two rectangles is given by : A = ¹/₁₈ (2x² – 40x + 400).
b. For what value of x is this area a minimum?
a. the first piece is xcm then the second piece must be (20 – x)cm

so this makes the perimeters of the first rectangle x and the second rectangle 20 – x

For the first rectangle, 2(L + B) = x  and if L = 2B then 2(2B + B) = x

so x = 6B

hence B = ^x/₆ and hence L = 2 × B = ^x/₃

so it follows that the area A₁ = LB = ^x/₃ × ^x/₆

$$=\frac{x^2}{18}$$

second rectangle: 20 – x = 2(L + B) and if L = 2B then 2(2B + B) = 20 – x

which means 6B = 20 – x

so B = ^{20 – x}/₆   and L = ^{20 – x}/₃

so it follows that area A₂ = LB

$$=\frac{20-x}{3}\times\frac{20-x}{6}$$

$$=\frac{400-40x+x^2}{18}$$

∴ total area A = ¹/₁₈(x² + 400 – 40x + x²)

A = ¹/₁₈(2x² – 40x + 400)

 

b. to find minimum need to differentiate and set A’ = 0

A’ = ¹/₁₈(4x – 40) = 0   solve for x

4x – 40 = 0

4x = 40

x = 10

A” = ¹/₁₈(2)   so when x = 10  A” > 0 ∴ concave up ∴ minimum

hence minimum area when x = 10cm

[Admin]

in reply to: Stationary Points, Points Of Inflection, Increasing, Decreasing, Concavity #9896

For the function for x³ – 6x² + 9x – 5, state the values of x, for which ^dy/_dx < 0 and ^d²y/_dx² < 0.

^dy/_dx = 3x² – 12x + 9 < 0
3(x² – 4x + 3) < 0
3(x – 3)(x – 1) < 0

1 < x < 3
^d²y/_dx² = 6x – 12 < 0
6x < 12

x < 2
combining that to satisfy both conditions: 1 < x < 2

[Admin]

in reply to: Fractions #9812

Simplify $$\frac{5c}{a^2+ab}-\frac{c}{a+b}$$ 

$$=\frac{5c}{a(a+b)}-\frac{c}{a+b}$$

$$=\frac{5c-c(a)}{a(a+b)}$$

$$=\frac{5c-ac}{a(a+b)}$$

$$=\frac{c(5-a)}{a(a+b)}$$

[Admin]

in reply to: Fractions #9811

Simplify $$x-y-\frac{x^2}{x+y}$$
$$=\frac{x(x+y)}{x+y}-\frac{y(x+y)}{x+y}-\frac{x^2}{x+y}$$

$$=\frac{x^2+xy-xy-y^2-x^2}{(x+y)}$$

$$=\frac{-y^2}{(x+y)}$$

[Admin]

in reply to: Quadratic Factors #9792

Factorise x⁴ – 3x² – 4
this is a hidden quadratic, probably the easiest way to do it is like this

let y = x², and now we rewrite

y² – 3y – 4 since (x²)² = x⁴

Factorise normally now: ie what multiplies to -4 and adds to -3: -4, +1

(y – 4)(y + 1)

then replace the y with x² so you have

(x² – 4)(x² + 1) then we factor the first bracket since is difference of two squares, ending up with

= (x – 2)(x + 2)(x² + 1)

[Admin]

in reply to: Quadratic Factors #9791

Factorise x² + 12x – 64.
= (x + 16)(x – 4)       16 × -4 = -64 and 16 – 4 = 12

[Admin]

in reply to: Equations reducible to quadratics #9757

Solve $${\left( {x^2 + \frac{1}{x^2}} \right)^2} – 9\left( {x^2 + \frac{1}{x^2}} \right) + 20 = 0$$ correct to 2 decimal places.

let $$y=x^2+\frac{1}{x^2}$$

y² – 9y + 20 = 0

(y – 5)(y – 4) = 0

y = 5, y = 4

$$x^2+\frac{1}{x^2}=5$$           $$x^2+\frac{1}{x^2}=4$$                     multiply throughout by x²

x⁴ + 1 = 5x²     x⁴ + 1 = 4x²

x⁴ – 5x² + 1  = 0     x⁴ – 4x² + 1  = 0      these are just another ‘hidden quadratic’

let a = x²

a² – 5a + 1 = 0        a² – 4a + 1 = 0

$$a=\frac{5\pm\sqrt{(-5)^2-4\times1\times1}}{2\times1}$$     $$a=\frac{4\pm\sqrt{(-4)^2-4\times1\times1}}{2\times1}$$

a = 0.2087      a = 4.79129       a = 0.2679     a = 3.732

x² = 0.2087       x² = 4.79129        x² = 0.2679     x² = 3.732     take the square root to find x, not forgetting the ±
x = ±0.46         x = ±2.19                 x = ±0.52        x = ±1.93

 

[Admin]

in reply to: Expanding #9511

Expand 2x(x + 3) – 3x(x – 4).

= 2x² + 6x – 3x² + 12x

= -x² + 18x

[Admin]

in reply to: Complementary Ratios #8142

Simplify: sec x – cosec (90° – x).

from our complimentary ratios we know that cosec (90° – x) = sec x
sec x – cosec (90° – x) = sec x – sec x
= 0

[Admin]

in reply to: Inequalities #8131

Solve $$\frac{q-2}{3}<2+\frac{3q}{4}$$

$$^{\times\cancel{3}\times4}\frac{q-2}{\cancel{3}}<2^{\times3\times4}+\frac{3q}{\cancel{4}}^{\times3\times\cancel{4}}$$

4q – 8 < 24 + 9q

-5q < 32

$$q>\frac{32}{-5}$$

$$q>-6\frac{2}{5}$$

[Author]

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