Quadratics

Solve `frac{3}{x+2}+frac{1}{x-1}=frac{1}{(x+1)(x+2)}`

Multiply both sides by `(x+2)(x-1)(x+1)`

`3(x-1)(x+1)+(x+2)(x+1)=1(x-1)`

Distribute the factors

`3x^2-3+x^2+3x+2=x-1`

Write in the form `ax^2+bx+c=0`

`4x^2+2x=0`

Factorise the common factors

`2x(2x+1)=0`

If  `pq=0`, then `p=0` or `q=0`

`2x=0` or `2x+1=0`

`x=0` or `x=-frac{1}{2}`

Validate x

`(x+2)(x-1)(x+1)!=0` 

`x+2!=0` or `x-1!=0` or `x+1!=0`

`x!=-2` or `x!=1` or `x!=-1`

Final solution

`x=0` or `x=-frac{1}{2}`

Solve `frac{x^2-2x-8}{x^2+7x+10}=0`

Multiply both sides by `x^2+7x+10`

`x^2-2x-8=0`

The two numbers that multiply to –8 `(axxc=-8)` and add up to –2 `(b=-2)` are –4 and 2

Rewrite

`x^2-4x+2x-8=0`

Factorise the common factors

`(x-4)(x+2)=0`

`x-4=0` or `x+2=0`

`x=4` or `x=-2`

Validate x

`x^2+7x+10!=0`

`(x+5)(x+2)!=0`

`x+5!=0` or `x+2!=0`

`x!=-5` or `x!=-2`

Final solution

`x=4`

Solve `(x^2-3x+1)^((2x^2+x-6))=1`

There are three cases in which this equation can be satisfied

● `x^2-3x+1=1`

Write in the form `ax^2+bx+c=0`

`x^2-3x=0`

Factorise the common factors

`x(x-3)=0`

`x=0` or `x-3=0`

`x=0` or `x=3`

● `2x^2+x-6=0`

The two numbers that multiply to –12 `(axxc=-12)` and add up to 1 `(b=1)` are 4 and –3

Rewrite

`2x^2+4x-3x-6=0`

Factorise the common factors

`(2x-3)(x+2)=0`

`2x-3=0` or `x+2=0`

`x=frac{3}{2}` or `x=-2`

● `x^2-3x+1=-1` and `2x^2+x-6` is even

For `x^2-3x+1=-1`, write in the form `ax^2+bx+c=0`

`x^2-3x+2=0`

The two numbers that multiply to 2 `(axxc=2)` and add up to –3 `(b=-3)` are –1 and –2

Rewrite

`x^2-x-2x+2=0`

Factorise the common factors

`(x-1)(x-2)=0`

`x-1=0` or `x-2=0`

`x=1` or `x=2`

Substitute x in the exponent `2x^2+x-6` 

For `x=1`

`2(1)^2+1-6=-3` (odd number)

For `x=2`

`2(2)^2+2-6=4` (even number)

Final solution

`x=2`

Solve `3^((2x^2+9x+2))=frac{1}{9}`

Rewrite 

`3^((2x^2+9x+2))=3^-2`

Equating powers of 3

`2x^2+9x+2=-2`

`2x^2+9x+4=0`

Factorise

`2x^2+8x+x+4=0`

`(2x+1)(x+4)=0`

`2x+1=0` or `x+4=0`

`x=-frac{1}{2}` or `x=-4`

The diagram shows a right-angled triangle with sides `2x` cm, `(2x+1)` cm and 29 cm.

a. Show that `2x^2+x-210=0`.

b. Find the lengths of the sides of the triangle.

a. Using Pythagoras

`(2x)^2+(2x+1)^2=29^2`

`4x^2+4x^2+4x+1=841`

`8x^2+4x-840=0`

Divide both sides by 4

`2x^2+x-210=0`

b. Factorise

`2x^2+21x-20x-210=0`

`(x-10)(2x+21)=0`

`x=10` or `2x+21=0`

`x=10` or `x=-frac{21}{2}=-10.5`

Substitute x values

For `x=10`

`(2xx10)^2+(2xx10+1)^2=29^2`

`400+441=841`

`841=841`

For `x=-10.5`

`(2xx10)^2+(2xx10+1)^2=29^2`

`441+400=841`

`841=841`

Final solution

`x=10` or `x=-10.5`

The area of the trapezium is 35.75 cm².

Find the value of x.

Area of a trapezium is `frac{1}{2}(a+b)h`

`frac{1}{2}[(x-1)+(x+3)]x=35.75`

Multiply both sides by 4

`2[(x-1)+(x+3)]x=143`

`2(2x+2)x=143`

`4x^2+4x-143=0`

Factorise

`4x^2+26x-22x-143=0`

`(2x-11)(2x+13)=0`

`2x-11=0` or `2x+13=0`

`x=5.5` or `x=-6.5`

x is the length of one side of the trapezium

Final solution

`x=5.5`

Express the following in the form `(x+a)^2+b`

`x^2-3x+4`

Find the square of half the coefficient –3

`-3-:2=-frac{3}{2}`

`(-frac{3}{2})^2=frac{9}{4}`

Add and subtract to the expression

`x^2-3x+frac{9}{4}-frac{9}{4}+4`

`(x^2-3x+frac{9}{4})-frac{9}{4}+4`

Write in the form of a perfect square

`(x-frac{3}{2})^2+frac{7}{4}`

Express the following in the form `a(x+b)^2+c`

`2x^2+7x+5`

Factorise

`2(x^2+frac{7}{2}x)+5`

Find the square of half the coefficient `frac{7}{2}`

`frac{7}{2}-:2=frac{7}{4}`

`(frac{7}{4})^2=frac{49}{16}`

Add and subtract to the expression

`2(x^2+frac{7}{2}x+frac{49}{16}-frac{49}{16})+5`

`2(x^2+frac{7}{2}x+frac{49}{16})+5-frac{49}{8}`

Write in the form of a perfect square

`2(x+frac{7}{4})^2-frac{9}{8}`

Express the following in the form `a-(x+b)^2`

`4-3x-x^2`

Rewrite

`4-(3x+x^2)`

Find the square of half the coefficient 3

`3-:2=frac{3}{2}`

`(frac{3}{2})^2=frac{9}{4}`

Add and subtract to the expression

`4-(3x+x^2+frac{9}{4}-frac{9}{4})`

`4+frac{9}{4}-(3x+x^2+frac{9}{4})`

Write in the form of a perfect square

`frac{25}{4}-(x+frac{3}{2})^2`

Express the following in the form `p-q(x+r)^2`

`2+5x-3x^2`

Factorise

`2-3(x^2-frac{5}{3}x)`

Find the square of half the coefficient `-frac{5}{3}`

`-frac{5}[3}-:2=-frac{5}{6}`

`(-frac{5}{6})^2=frac{25}{36}`

Add and subtract to the expression

`2-3(x^2-frac{5}{3}x+frac{25}{36}-frac{25}{36})`

`2-3(x^2-frac{5}{3}x+frac{25}{36})+frac{25}{12}`

Write in the form of a perfect square

`frac{49}{12}-3(x-frac{5}{6})^2`

Express the following in the form `(ax+b)^2+c`

`25x^2+40x-4`

Using an algebraic method

`(ax+b)^2+c=a^2x^2+2abx+b^2+c`

Then

`a^2=25`, `2ab=40`, `b^2+c=-4`

For `a^2=25`

`a=+-5`

● If `a=5`

`10b=40`, so `b=4`

`4^2+c=-4`, so `c=-20`

● If `a=-5`

`-10b=40`, so `b=-4`

`(-4)^2+c=-4`, so `c=-20`

Final solution

`25x^2+40x-4=(5x+4)^2-20=(-5x-4)^2-20`

Solve by completing the square `2x^2-8x-3=0`

Leave the answers in surd form

Factorise

`2(x^2-4x-frac{3}{2})=0`

`x^2-4x-frac{3}{2}=0`

The square of half the coefficient –4

`-4-:2=-2`

`(-2)^2=4`

Add and subtract to the expression

`x^2-4x-frac{3}{2}+4-4=0`

`x^2-4x+4=frac{3}{2}+4`

Write in the form of a perfect square

`(x-2)^2=frac{11}{2}`

Square root both sides

`x-2=+-sqrt{frac{11}{2}}`

`x=frac{4+-sqrt{22}}{2}`

Solve `frac{5}{x+2}+frac{3}{x-4}=2`

Leave the answers in surd form

Multiply both sides by `(x+2)(x-4)`

`5(x-4)+3(x+2)=2(x+2)(x-4)`

Distribute the factors

`5x-20+3x+6=2x^2-8x+4x-16`

Write in the form `ax^2+bx+c=0`

`2x^2-12x-2=0`

Factorise

`2(x^2-6x-1)=0`

`x^2-6x-1=0`

The square of half the coefficient –6

`-6-:2=-3`

`(-3)^2=9`

Add and subtract to the expression

`x^2-6x-1+9-9=0`

`x^2-6x+9=10`

Write in the form of a perfect square

`(x-3)^2=10`

Square root both sides

`x-3=+-sqrt10`

`x=3+-sqrt10`

The diagram shows a right-angled triangle with sides `x` m, `(2x+5)` m and 10 m. 

Find the value of x. Leave the answers in surd form.

Using Pythagoras

`x^2+(2x+5)^2=10^2` 

 `x^2+4x^2+20x+25=100`

`5x^2+20x-75=0`

Factorise

`5(x^2+4x-15)=0`

 `x^2+4x-15=0`

 The square of half the coefficient 4

`4-:2=2`

`(2)^2=4`

 Add and subtract to the expression

`x^2+4x-15+4-4=0`

`x^2+4x+4=15+4`

Write in the form of a perfect square

`(x+2)^2=19`

Square root both sides

 `x+2=+-sqrt(19)`

 `x=-2+sqrt(19)` or `x=-2-sqrt(19)`

 x is the length of one side of the triangle

Final solution

  `x=-2+sqrt(19)`

Find the real solutions of the equation `(3x^2+5x-7)^4=1`

`a^4=1` (exponent is an even number), then `a=1`  or `a=-1`

`3x^2+5x-7=1` or `3x^2+5x-7=-1`

For `3x^2+5x-7=1`

`3x^2+5x-8=0`

Factorise

`3(x^2+frac{5}{3}x-frac{8}{3})=0`

`x^2+frac{5}{3}x-frac{8}{3}=0`

The square of half the coefficient `frac{5}{3}`

`frac{5}{3}-:2=frac{5}{6}`

`(frac{5}{6})^2=frac{25}{36}`

Add and subtract to the expression

`x^2+frac{5}{3}x-frac{8}{3}+frac{25}{36}-frac{25}{36}=0`

`x^2+frac{5}{3}x+frac{25}{36}=frac{25}{36}+frac{8}{3}`

Write in the form of a perfect square

`(x+frac{5}{6})^2=frac{121}{36}`

Square root both sides

`x+frac{5}{6}=+-frac{11}{6}`

`x+frac{5}{6}=frac{11}{6}` or `x+frac{5}{6}=-frac{11}{6}`

`x=1` or `x=-frac{8}{3}`

● For `3x^2+5x-7=-1`

`3x^2+5x-6=0`

Factorise

`3(x^2+frac{5}{3}x-2)=0`

`x^2+frac{5}{3}x-2=0`

The square of half the coefficient `frac{5}{3}`

`frac{5}{3}-:2=frac{5}{6}`

`(frac{5}{6})^2=frac{25}{36}`

Add and subtract to the expression

`x^2+frac{5}{3}x-2+frac{25}{36}-frac{25}{36}=0`

`x^2+frac{5}{3}x+frac{25}{36}=frac{25}{36}+2`

Write in the form of a perfect square

`(x+frac{5}{6})^2=frac{97}{36}`

Square root both sides

`x+frac{5}{6}=+-frac{sqrt{97}}{6}`

`x+frac{5}{6}=frac{sqrt{97}}{6}` or `x+frac{5}{6}=-frac{sqrt{97}}{6}`

`x=frac{-5+sqrt{97}}{6}` or `x=frac{-5-sqrt{97}}{6}`

a. Express `x^2+6x+2` in the form `(x+a)^2+b`, where a and b are constants.

b.  Hence, or otherwise, find the set of values x of for which `x^2+6x+2>9`

a. Find the square of half the coefficient 6

 `6-:2=3`

 `(3)^2=9`

Add and subtract to the expression

 `x^2+6x+9-9+2`

Write in the form `(x+a)^2+b`

 `(x+3)^2-7`

Where `a=3` and `b=-7`

b. From the completed square form

 `(x+3)^2-7>9`

 `(x+3)^2>16`

Square root both sides

 `|x+3|>4`

`x+3>4` or `x+3<-4`

`x>1` or `x<-7`

Showing all necessary working, solve the equation `4x-11x^frac{1}{2}+6=0`

Substitute `u=x^frac{1}{2}`

`u^2=x`

Then

`4u^2-11u+6=0`

The two numbers that multiply to 24 `(axxc=24)` and add up to –11 `(b=-11)` are –8  and –3

Rewrite

`4u^2-8u-3u+6=0`

Factorise the common factors

`(4u-3)(u-2)=0`

`4u-3=0` or `u-2=0`

`u=frac{3}{4}` or `u=2`

`x^frac{1}{2}=frac{3}{4}` or `x^frac{1}{2}=2`

`(x^frac{1}{2})^2=frac{3^2}{4^2}` or `(x^frac{1}{2})^2=2^2`

`x=frac{9}{16}` or `x=4`

The equation of a curve is `y=(2k-3)x^2-kx-(k-2)`, where k is a constant. The line `y=3x-4` is a tangent to the curve.

Find the value of k.

Equate the curve and the line

`(2k-3)x^2-kx-(k-2)=3x-4`

Write in the form `ax^2+bx+c=0`

 `(2k-3)x^2-(k+3)x-(k-6)=0`

Since the line is a tangent (one point of intersection), the discriminant  `b^2-4ac`of the quadratic must be zero

 `(-(k+3))^2-4(2k-3)(-(k-6))=0`

`(k+3)^2+4(2k-3)(k-6)=0`

`k^2+6k+9+8k^2-48k-12k+72=0`

`9k^2-54k+81=0`

Using quadratic formula `k=frac{-b+-sqrt{b^2-4ac}){2a}`

`k=frac{54+-sqrt{(-54)^2-4(9)(81)}){2(9)}`

`k=frac{54+-sqrt{2916-2916}){18}`

`k=frac{54}{18}`

`k=3`

The equation of a line is `y=mx+c`, where m and c are constants, and the equation of a curve is `xy=16`.

a. Given that the line is a tangent to the curve, express m in terms of c.

b. Given instead that `m=-4`, find the set of values of c for which the line intersects the curve at two distinct points.

a. Substitute y

 `x(mx+c)=16`

 `mx^2+cx-16=0`

Since the line is a tangent (one point of intersection), the discriminant `b^2-4ac` of the quadratic must be zero

`Delta=b^2-4ac=0`

 `c^2-4m(-16)=0`

 `c^2=-64m`

 `m=frac{-c^2}{64}`

b. Since there are two distinct points of intersection, the discriminant `b^2-4ac` of the quadratic must be greater than zero

 `Delta=b^2-4ac>0`

Substitute `m=-4`

 `c^2-4(-4)(-16)>0`

 `c^2>256`

Square root both sides

 `|c|>16`

`c>16` and `c<-16`

The line with equation `y=kx-k`, where k is a positive constant, is a tangent to the curve with equation `y=-frac{1}{2x}`.

Find, in either order, the value of k and the coordinates of the point where the tangent meets the curve.

Equate the curve and the line

`kx-k=-frac{1}{2x}`

Multiply both sides by `2x`

 `2kx^2-2kx=-1`

Write in the form `ax^2+bx+c=0`

 `2kx^2-2kx+1=0`

Since the line is a tangent (one point of intersection), the discriminant `b^2-4ac` of the quadratic must be zero

 `Delta=b^2-4ac=0`

 `(-(2k))^2-4(2k)(1)=0`

 `4k^2-8k=0`

 `4k(k-2)=0`

`4k=0` or `k-2=0`

`k=0` or `k=2`

Final solution

`k=2` (k is a positive constant)