1.3 Complex Quadratic Equations

1. Complex Quadratics

Many equations that contain higher powers or roots can be transformed into a standard quadratic form $a u^{2} + b u + c = 0$ using an appropriate substitution.

Common Substitutions

$x^{4} + k x^{2} + c = 0$ → Let $u = x^{2}$
$x^{6} + k x^{3} + c = 0$ → Let $u = x^{3}$
$x + k \sqrt{x} + c = 0$ → Let $u = \sqrt{x}$
$a^{2 x} + k (a^{x}) + c = 0$ → Let $u = a^{x}$

Validity Check

Always check if the value of $u$ is valid for the substitution used.

If $u = x^{2}$ , $u$ must be $\geq 0$ .
If $u = \sqrt{x}$ , $u$ must be $\geq 0$ .
If $u = a^{x}$ , $u$ must be $> 0$ .

Reject any solutions for $u$ that violate these conditions before solving for $x$ .

Worked Examples

1. Find the real values of

x

for

x^{4} - 10 x^{2} + 9 = 0

Recognize 'hidden quadratic' form:

(x^{2})^{2} - 10 (x^{2}) + 9 = 0

Let

u = x^{2}

. Equation becomes

u^{2} - 10 u + 9 = 0

Factorise:

(u - 9) (u - 1) = 0

u = 9

u = 1

Substitute

x^{2}

back:

x^{2} = 9 \Rightarrow x = \pm 3

x^{2} = 1 \Rightarrow x = \pm 1

Solutions:

x = \pm 1, \pm 3

2. Find the real values of

x

for

x^{6} - 5 x^{3} - 6 = 0

Let

u = x^{3}

. Then

u^{2} = x^{6}

Equation:

u^{2} - 5 u - 6 = 0

Factorise:

(u - 6) (u + 1) = 0

u = 6

u = - 1

Substitute back:

x^{3} = 6 \Rightarrow x = \sqrt[3]{6}

x^{3} = - 1 \Rightarrow x = - 1

Solutions:

x = \sqrt[3]{6}, - 1

3. Find the real values of

x

for

x^{4} - 7 x^{2} + 10 = 0

Let

u = x^{2}

. Equation:

u^{2} - 7 u + 10 = 0

Factorise:

(u - 5) (u - 2) = 0

u = 5

u = 2

x^{2} = 5 \Rightarrow x = \pm \sqrt{5}

x^{2} = 2 \Rightarrow x = \pm \sqrt{2}

Solutions:

x = \pm \sqrt{5}, \pm \sqrt{2}

4. Find the real values of

x

for

3 x^{4} - 11 x^{2} + 2 = 0

Let

u = x^{2}

. Equation:

3 u^{2} - 11 u + 2 = 0

Use quadratic formula (or factorise if possible? A=3, C=2, AC=6. Sum=-11. No integer factors).

Formula:

u = \frac{11 \pm \sqrt{121 - 24}}{6} = \frac{11 \pm \sqrt{97}}{6}

x^{2} = \frac{11 + \sqrt{97}}{6}

x^{2} = \frac{11 - \sqrt{97}}{6}

Both values are positive (

\sqrt{97} \approx 9.8

11 - 9.8 > 0

), so all roots are real.

Solutions:

x = \pm \sqrt{\frac{11 + \sqrt{97}}{6}}, \pm \sqrt{\frac{11 - \sqrt{97}}{6}}

5. Solve

3 x - 4 \sqrt{x} + 1 = 0

Let

u = \sqrt{x}

(where

u \geq 0

). Then

x = u^{2}

Equation:

3 u^{2} - 4 u + 1 = 0

Factorise:

(3 u - 1) (u - 1) = 0

u = 1 / 3

u = 1

Substitute back:

\sqrt{x} = 1 / 3 \Rightarrow x = 1 / 9

\sqrt{x} = 1 \Rightarrow x = 1

Solutions:

x = 1 / 9, 1

6. Solve

\sqrt{x} (\sqrt{x} - 2) = 3

Expand:

x - 2 \sqrt{x} = 3

Rearrange:

x - 2 \sqrt{x} - 3 = 0

Let

u = \sqrt{x}

u^{2} - 2 u - 3 = 0

Factorise:

(u - 3) (u + 1) = 0

u = 3

u = - 1

Validity Check:

\sqrt{x}

cannot be negative.

u = - 1

is rejected.

Only

u = 3 \Rightarrow x = 9

Solution:

x = 9

7. Solve

5 x - 12 \sqrt{x} + 4 = 0

Let

u = \sqrt{x}

5 u^{2} - 12 u + 4 = 0

Factorise:

(5 u - 2) (u - 2) = 0

u = 2 / 5

u = 2

\sqrt{x} = 2 / 5 \Rightarrow x = 4 / 25 = 0.16

\sqrt{x} = 2 \Rightarrow x = 4

Solutions:

x = 0.16, 4

8. Solve

6 x - 5 \sqrt{x} - 1 = 0

Let

u = \sqrt{x}

6 u^{2} - 5 u - 1 = 0

Factorise:

(6 u + 1) (u - 1) = 0

u = 1

u = - 1 / 6

u

must be positive, so reject

- 1 / 6

\sqrt{x} = 1 \Rightarrow x = 1

Solution:

x = 1

9. Curve

y = 4 \sqrt{x}

and line

y = x + 2

intersect at A, B. Find coords.

Equate:

x + 2 = 4 \sqrt{x}

Rearrange:

x - 4 \sqrt{x} + 2 = 0

Let

u = \sqrt{x}

u^{2} - 4 u + 2 = 0

Formula:

u = \frac{4 \pm \sqrt{16 - 8}}{2} = 2 \pm \sqrt{2}

Find

x

x_{1} = (2 + \sqrt{2})^{2} = 4 + 2 + 4 \sqrt{2} = 6 + 4 \sqrt{2}

x_{2} = (2 - \sqrt{2})^{2} = 6 - 4 \sqrt{2}

Find

y

y = x + 2

y_{1} = 8 + 4 \sqrt{2}

y_{2} = 8 - 4 \sqrt{2}

Coords:

(6 + 4 \sqrt{2}, 8 + 4 \sqrt{2})

and

(6 - 4 \sqrt{2}, 8 - 4 \sqrt{2})

10. Find real

x

for

x^{8} - 13 x^{4} + 36 = 0

Let

u = x^{4}

. Equation:

u^{2} - 13 u + 36 = 0

Factorise:

(u - 9) (u - 4) = 0

u = 9

u = 4

x^{4} = 9 \Rightarrow x^{2} = 3

(since real)

\Rightarrow x = \pm \sqrt{3}

x^{4} = 4 \Rightarrow x^{2} = 2

(since real)

\Rightarrow x = \pm \sqrt{2}

Solutions:

x = \pm \sqrt{2}, \pm \sqrt{3}

2. Geometric Applications

Solving intersections between lines and curves, especially those involving $\sqrt{x}$ or unknown coefficients, requires careful algebraic manipulation.

Key Strategies

Isolating the Radical: When solving $A \sqrt{x} = B x + C$ , isolate the square root term before squaring both sides. Note that squaring can introduce extraneous solutions, so check your answers!
Unknown Coefficients: If a curve passes through a point $(x_{1}, y_{1})$ , substitute these values into the equation to create a linear system to solve for $a, b, c$ .
Distance Between Points: Use the formula $d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ or simplified versions using gradients for lines.

Equation of chord intersecting curve

Worked Examples: Geometry & Graphs

1. Curve

y = 3 \sqrt{x}

and line

2 y = x + 4

intersect at A, B. Find A and B and length AB.

Rearrange line:

y = \frac{1}{2} x + 2

Equate:

3 \sqrt{x} = \frac{1}{2} x + 2

Multiply by 2:

6 \sqrt{x} = x + 4

Rearrange to form quadratic in

\sqrt{x}

x - 6 \sqrt{x} + 4 = 0

Let

u = \sqrt{x}

u^{2} - 6 u + 4 = 0

Use formula:

u = \frac{6 \pm \sqrt{36 - 16}}{2}

= \frac{6 \pm \sqrt{20}}{2} = 3 \pm \sqrt{5}

u_{1} = 3 + \sqrt{5} \approx 5.236

u_{2} = 3 - \sqrt{5} \approx 0.764

(Both positive, so valid).

Find

x

x_{1} = (3 + \sqrt{5})^{2}

= 9 + 5 + 6 \sqrt{5}

= 14 + 6 \sqrt{5}

x_{2} = (3 - \sqrt{5})^{2} = 14 - 6 \sqrt{5}

Find

y

y = 3 \sqrt{x} = 3 u

y_{1} = 3 (3 + \sqrt{5}) = 9 + 3 \sqrt{5}

y_{2} = 3 (3 - \sqrt{5}) = 9 - 3 \sqrt{5}

Coords:

(14 + 6 \sqrt{5}, 9 + 3 \sqrt{5})

(14 - 6 \sqrt{5}, 9 - 3 \sqrt{5})

Distance AB:

\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

Δ x = 12 \sqrt{5}

Δ y = 6 \sqrt{5}

A B = \sqrt{(12 \sqrt{5})^{2} + (6 \sqrt{5})^{2}}

= \sqrt{720 + 180}

= \sqrt{900} = 30

2. Find

a, b, c

for

y = a x + b \sqrt{x} + c

passing through

(1, 0), (49 / 4, 0), (0, 7)

Slope-intercept form for

x = 0

y = a (0) + b (0) + c = 7

. So,

c = 7

Passes through

(1, 0)

0 = a (1) + b (1) + 7 \Rightarrow a + b = - 7

. (Eq 1)

Passes through

(12.25, 0)

0 = 12.25 a + 3.5 b + 7

. (Note

\sqrt{49 / 4} = 7 / 2 = 3.5

Multiply by 4 to clear decimals/fractions:

0 = 49 a + 14 b + 28

Divide by 7:

7 a + 2 b + 4 = 0 \Rightarrow 7 a + 2 b = - 4

. (Eq 2)

From Eq 1:

b = - 7 - a

. Substitute into Eq 2:

7 a + 2 (- 7 - a) = - 4

7 a - 14 - 2 a = - 4 \Rightarrow 5 a = 10 \Rightarrow a = 2

Find

b

b = - 7 - 2 = - 9

Result:

a = 2, b = - 9, c = 7

3. Find real

x

for

x^{4} - 4 x^{3} + 5 x^{2} - 4 x + 4 = 0

Notice the coefficients

1, - 4, 5, - 4, 1

are almost symmetric.

Try small integers. Let

x = 2

Substitute

x = 2

16 - 4 (8) + 5 (4) - 4 (2) + 4

= 16 - 32 + 20 - 8 + 4 = 0

(x - 2)

is a factor.

Perform division or inspection:

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

Group second bracket:

x^{2} (x - 2) + 1 (x - 2) = (x^{2} + 1) (x - 2)

Full equation:

(x - 2) (x - 2) (x^{2} + 1)

= (x - 2)^{2} (x^{2} + 1) = 0

Real solutions come from

x - 2 = 0

x^{2} + 1 = 0

has no real roots.

Result:

x = 2

4. Find

a, b, c

for

y = a x + b \sqrt{x} + c

cutting x-axis at

1, 25

and y-axis at

4

y-intercept

(0, 4) \Rightarrow c = 4

x = 1, y = 0

a + b + 4 = 0 \Rightarrow a + b = - 4

x = 25, y = 0

25 a + 5 b + 4 = 0

From first eq:

b = - 4 - a

. Substitute:

25 a + 5 (- 4 - a) + 4 = 0

25 a - 20 - 5 a + 4 = 0

20 a = 16 \Rightarrow a = 16 / 20 = 0.8

b = - 4 - 0.8 = - 4.8

Result:

a = 0.8, b = - 4.8, c = 4

5. Solve

\frac{6}{\sqrt{x}} + \sqrt{x} = 7

Multiply by

\sqrt{x}

6 + x = 7 \sqrt{x}

Rearrange:

x - 7 \sqrt{x} + 6 = 0

Let

u = \sqrt{x}

u^{2} - 7 u + 6 = 0

Factorise:

(u - 6) (u - 1) = 0

u = 6

u = 1

\sqrt{x} = 6 \Rightarrow x = 36

\sqrt{x} = 1 \Rightarrow x = 1

Solutions:

x = 1, 36

3. Mixed Questions

Additional questions modeled after Cambridge A-Level examinations.

1. Solve the equation

4 x^{4} + 3 x^{2} - 1 = 0

Let

u = x^{2}

4 u^{2} + 3 u - 1 = 0

Factorise:

(4 u - 1) (u + 1) = 0

u = 1 / 4

u = - 1

Since

x^{2} = u

u

must be non-negative for real roots.

Reject

u = - 1

x^{2} = 1 / 4 \Rightarrow x = \pm 1 / 2

Solutions:

x = 0.5, - 0.5

2. Find the coordinates of the points of intersection of the line

y = 2 x + 1

and the curve

y = \frac{3}{x} + 2

Equate:

2 x + 1 = \frac{3}{x} + 2

Simplify:

2 x - 1 = \frac{3}{x}

Multiply by

x

2 x^{2} - x = 3

2 x^{2} - x - 3 = 0

Factorise:

(2 x - 3) (x + 1) = 0

x = 1.5

x = - 1

Find

y

x = 1.5 \Rightarrow y = 2 (1.5) + 1 = 4

x = - 1 \Rightarrow y = 2 (- 1) + 1 = - 1

Points:

(1.5, 4)

and

(- 1, - 1)

3. Solve

3 (2^{2 x}) - 10 (2^{x}) + 3 = 0

Let

u = 2^{x}

Recall

2^{2 x} = (2^{x})^{2} = u^{2}

Equation:

3 u^{2} - 10 u + 3 = 0

Factorise:

(3 u - 1) (u - 3) = 0

u = 1 / 3

u = 3

2^{x} = 1 / 3 \Rightarrow x = \log_{2} (1 / 3) = - \log_{2} (3)

2^{x} = 3 \Rightarrow x = \log_{2} (3)

Exact answers:

\pm \log_{2} 3

(or using

\ln

\frac{\pm \ln 3}{\ln 2}

4. Solve the equation

2 x - 7 \sqrt{x} + 3 = 0

Let

u = \sqrt{x}

. Equation:

2 u^{2} - 7 u + 3 = 0

Factorise:

(2 u - 1) (u - 3) = 0

u = 1 / 2

u = 3

\sqrt{x} = 1 / 2 \Rightarrow x = 1 / 4

\sqrt{x} = 3 \Rightarrow x = 9

Solutions:

x = 0.25, 9

5. The line

y = k x - 2

is a tangent to the curve

y = x^{2} - 4 x + 7

. Find the possible values of

k

Equate line and curve:

x^{2} - 4 x + 7 = k x - 2

Rearrange:

x^{2} - (4 + k) x + 9 = 0

For a tangent, the discriminant

b^{2} - 4 a c = 0

(- (4 + k))^{2} - 4 (1) (9) = 0

(4 + k)^{2} - 36 = 0

(4 + k)^{2} = 36 \Rightarrow 4 + k = \pm 6

Case 1:

4 + k = 6 \Rightarrow k = 2

Case 2:

4 + k = - 6 \Rightarrow k = - 10

Values:

k = 2, - 10

6. Find the real roots of the equation

x^{6} + 7 x^{3} - 8 = 0

Let

u = x^{3}

. Equation:

u^{2} + 7 u - 8 = 0

Factorise:

(u + 8) (u - 1) = 0

u = - 8

u = 1

x^{3} = - 8 \Rightarrow x = - 2

x^{3} = 1 \Rightarrow x = 1

Solutions:

x = - 2, 1