MathsISC Class 11 · Quadratic Equations

Equations Reducible to Quadratic, Common Roots, Sign and Extreme Values of a Quadratic

Q: What is an extraneous root?

A root of the transformed equation that does not satisfy the original equation.

Q: How do I find the common root?

Subtract the two equations (after equalising the x² coefficients) to get a linear equation in x.

Q: How can I find the range of a quadratic?

Use its extreme value: [min, ∞) for a > 0 or (−∞, max] for a < 0.

Updated 17 October 2026 · Fact-checked

Equations such as x⁴ − 13x² + 36 = 0 or √(x + 5) = x − 1 are reduced to quadratics by substitution or squaring, always checking the roots. Two quadratics have a common root when the root of their difference satisfies both. A quadratic ax² + bx + c has its extreme value (c − b²/4a) at x = −b/2a: a minimum for a > 0 and a maximum for a < 0.

Understand Equations Reducible to Quadratic, Common Roots and Extreme Values

Reducible equations. (1) *Substitution:* in $x^{4} - 13 x^{2} + 36 = 0$ put $y = x^{2}$ to get $y^{2} - 13 y + 36 = 0$ , then return to x. (2) *Reciprocal types:* $x + \frac{1}{x} = t$ . (3) *Radical equations:* isolate the root, square both sides, solve the resulting quadratic and check every root in the original equation, because squaring can introduce extraneous roots. (4) *Equations with powers:* $2^{2 x} - 5 \cdot 2^{x} + 4 = 0$ with $t = 2^{x}$ .

Common roots. For $a_{1} x^{2} + b_{1} x + c_{1} = 0$ and $a_{2} x^{2} + b_{2} x + c_{2} = 0$ with one common root α, subtract after making the coefficients of $x^{2}$ equal; solve the resulting linear equation for α, and substitute in either equation to find the unknown parameter. Both roots are common if the coefficients are proportional.

Sign of a quadratic. For $f (x) = a x^{2} + b x + c$ with $D < 0$ : f has the same sign as a for all real x. With $D > 0$ and roots p < q: f has the sign of a outside [p, q] and the opposite sign between the roots. With $D = 0$ : f has the sign of a except at the repeated root where it is zero.

Extreme values. Complete the square: $f (x) = a (x + \frac{b}{2 a})^{2} + \frac{4 a c - b ^{2}}{4 a}$ . For a > 0 the minimum value is $\frac{4 a c - b ^{2}}{4 a}$ at $x = - \frac{b}{2 a}$ (the graph opens upwards); for a < 0 it is the maximum value (the graph opens downwards). The range follows.

Equations Reducible to Quadratic, Common Roots and Extreme Values: formulas to remember

Substitution: $Biquadratic: $y = x^2$ | Exponential: $t = 2^x$ | Reciprocal: $t = x + 1/x$$; Convert to a quadratic.
Extreme value: $At $x = -\frac{b}{2a}$: $f = \frac{4ac - b^2}{4a}$$; Minimum if a > 0; maximum if a < 0.
Sign: $D < 0 : s i g n o f a e v er y w h er e ∣ D > 0 : s i g n o f a o u t s i d e t h er oo t s$; Opposite sign between the roots.
Common root: $S u b t r a c tt h ee q u a t i o n s t o f in d α, t h e n s u b s t i t u t e$; Check that α satisfies both.

Equations Reducible to Quadratic, Common Roots and Extreme Values in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Maximum profit of a stall

Suppose the weekly profit of a stall is $P = - 2 x^{2} + 8 x - 3$ thousand rupees when the price is x hundred rupees.

Find the price giving the maximum profit, the maximum profit, and the range of prices for which the stall makes a profit.

Try it first, then show the solutionSolution

Vertex: $x = - \frac{b}{2 a} = - \frac{8}{- 4} = 2$ ; $P (2) = - 8 + 16 - 3 = 5$ thousand rupees.
Profit when $P > 0$ : roots of $2 x^{2} - 8 x + 3 = 0$ are $x = 2 \pm \frac{10}{2} \approx 0.42, 3.58$ .
Profit is made for prices between about 0.42 and 3.58 hundred rupees.

Answer: Price 200 rupees; maximum profit 5000 rupees; profit for roughly 42 to 358 rupees.

Memory hook: The graph opens downwards, so the vertex is a maximum.

Tricks to remember Equations Reducible to Quadratic, Common Roots and Extreme Values

Short memory aids. Say each one aloud once, then use it on a question the same day.

Always check radical roots
Squaring can create fake solutions.
Use it when: Radical equations.
Vertex at −b/2a
The extreme value lies at the axis of symmetry.
Use it when: Maximum and minimum.

Equations Reducible to Quadratic, Common Roots and Extreme Values: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Minimum value · 1 mark · MCQBoard level

The minimum value of $x^{2} - 6 x + 11$ is:
(a) 2
(b) 3
(c) 11
(d) −2

Try it first, then show the solutionSolution

At $x = 3$ : $9 - 18 + 11 = 2$ .

Answer: (a) 2

Example 2: A biquadratic equation · 3 marks · Short answerBoard level

Solve $x^{4} - 13 x^{2} + 36 = 0$ .

Try it first, then show the solutionSolution

Put $y = x^{2}$ : $y^{2} - 13 y + 36 = 0$ . (1 mark)
$(y - 4) (y - 9) = 0$ , so $y = 4$ or $9$ . (1 mark)
$x^{2} = 4$ gives $x = \pm 2$ ; $x^{2} = 9$ gives $x = \pm 3$ . (1 mark)

Answer: x = ±2, ±3.

Example 3: A radical equation · 3 marks · Short answerHarder board level

Solve $x + 5 = x - 1$ and check the roots.

Try it first, then show the solutionSolution

Square: $x + 5 = x^{2} - 2 x + 1$ , i.e. $x^{2} - 3 x - 4 = 0$ . (1 mark)
$(x - 4) (x + 1) = 0$ , so $x = 4$ or $x = - 1$ . (1 mark)
Check: $x = 4$ : $9 = 3 = 4 - 1$ ✓. $x = - 1$ : $4 = 2 \neq = - 2$ ✗. So only $x = 4$ . (1 mark)

Answer: x = 4 (x = −1 is extraneous).

Example 4: Common root · 4 marks · Short answerHarder board level

Find the value of k for which the equations $x^{2} + x + k = 0$ and $x^{2} + k x + 1 = 0$ have a common root.

Try it first, then show the solutionSolution

Subtract: $(1 - k) x + (k - 1) = 0$ , i.e. $(1 - k) (x - 1) = 0$ . (1.5 marks)
If $k \neq = 1$ , the common root is $x = 1$ . (0.5 mark)
Substitute in the first: $1 + 1 + k = 0$ , so $k = - 2$ ; check the second: $1 - 2 + 1 = 0$ ✓. (1 mark)
If $k = 1$ the equations are identical ( $x^{2} + x + 1$ ), so both roots are common. (1 mark)

Answer: k = −2 (common root 1) or k = 1 (both roots common).

Example 5: Extreme value and sign · 5 marks · Long answerHOTS / competency

For $f (x) = - 2 x^{2} + 8 x - 3$ find the maximum value and where it occurs, the range of f, and the values of x for which f(x) > 0.

Try it first, then show the solutionSolution

The vertex: $x = - \frac{8}{2 ( - 2 )} = 2$ ; maximum $f (2) = - 8 + 16 - 3 = 5$ . (1.5 marks)
Range: $(- \infty, 5]$ . (1 mark)
$f (x) > 0$ when $2 x^{2} - 8 x + 3 < 0$ : roots $x = \frac{8 \pm 40}{4} = 2 \pm \frac{10}{2}$ . (1.5 marks)
So $f (x) > 0$ for $2 - 1.58 < x < 2 + 1.58$ , i.e. approximately $0.42 < x < 3.58$ . (1 mark)

Answer: Maximum 5 at x = 2; range (−∞, 5]; f > 0 for about 0.42 < x < 3.58.

Where students lose marks in Equations Reducible to Quadratic, Common Roots and Extreme Values

The slips examiners see most often, and what to write instead.

Mistake: Not checking radical roots
Accepting x = −1 for √(x + 5) = x − 1.
Do this: Substitute every root into the original equation.
Usually costs: 2 marks.
Mistake: Mixing the maximum and minimum
Saying a > 0 gives a maximum.
Do this: a > 0: minimum; a < 0: maximum.
Usually costs: 1 mark.
Mistake: Leaving y instead of returning to x
Stopping at the values of y.
Do this: Substitute back to find x.
Usually costs: 1 to 2 marks.

Exam tips

State the substitution and the new equation clearly.
Check all roots in the original equation.
Use the vertex x = −b/2a for max or min problems.
Sketch the parabola to find where it is positive or negative.

Practice questions on Equations Reducible to Quadratic, Common Roots and Extreme Values

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The equation $x^{4} - 5 x^{2} + 4 = 0$ has how many real roots?
(a) 2
(b) 4
(c) 0
(d) 1
Hint
y = x² gives 1 and 4.
Answer
(b) 4
Q2 · 2 marks · Short answerBoard level
Find the minimum value of $x^{2} + 4 x + 7$ .
Hint
Complete the square.
Answer
$(x + 2)^{2} + 3$ ; minimum 3 at $x = - 2$ .
Q3 · 3 marks · Short answerHarder board level
Solve $2^{2 x} - 5 \cdot 2^{x} + 4 = 0$ .
Hint
Put t = 2ˣ.
Answer
$t^{2} - 5 t + 4 = 0$ gives $t = 1$ or $4$ , so $x = 0$ or $2$ .
Q4 · 3 marks · Short answerHarder board level
Solve $x + \frac{1}{x} = \frac{5}{2}$ .
Hint
Multiply by 2x.
Answer
$2 x^{2} - 5 x + 2 = 0$ gives $x = 2$ or $\frac{1}{2}$ .
Q5 · 3 marks · Short answerHarder board level
Show that $x^{2} + x + 1$ is positive for every real x and find its least value.
Hint
D < 0 and a > 0.
Answer
$D = 1 - 4 = - 3 < 0$ and $a > 0$ , so it is always positive. Least value $= \frac{4 - 1}{4} = \frac{3}{4}$ at $x = - \frac{1}{2}$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a farmer wants to fence a rectangular field against a long straight wall using 40 m of fencing on three sides.
(i) Write the area in terms of the width x. (1 mark)
(ii) Find the width giving the maximum area. (1 mark)
(iii) Find the maximum area. (1 mark)
(iv) Find the length of the field. (1 mark)
Hint
Length = 40 − 2x.
Answer
(i) $A = x (40 - 2 x) = - 2 x^{2} + 40 x$ . (ii) $x = - \frac{40}{2 ( - 2 )} = 10$ m. (iii) $A = 400 - 200 = 200$ m². (iv) $40 - 20 = 20$ m.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Vertex of ax² + bx + c?Show answer
x = −b/2a.
2.Minimum when?Show answer
a > 0.
3.Extraneous roots arise from?Show answer
Squaring.
4.Sign of a quadratic with D < 0?Show answer
The sign of a.

Equations Reducible to Quadratic, Common Roots and Extreme Values: frequently asked questions

What is an extraneous root?

A root of the transformed equation that does not satisfy the original equation.

How do I find the common root?

Subtract the two equations (after equalising the x² coefficients) to get a linear equation in x.

How can I find the range of a quadratic?

Use its extreme value: [min, ∞) for a > 0 or (−∞, max] for a < 0.