MathsCBSE Class 10 · Quadratic Equations

Standard Form of a Quadratic Equation for Class 10

Updated 13 October 2026 · Fact-checked

A quadratic equation in x is any equation that can be rearranged as ax² + bx + c = 0, where a, b and c are real numbers and a is not zero. The board asks you to simplify an equation, decide whether it is quadratic, read a, b and c with their signs, and form equations from situations.

Read a, b and c from the standard form; b and c may be zero, but if a is zero the x² term vanishes and the equation is no longer quadratic.

Illustration drawn for QuestionBuddy.

Understand Standard Form of a Quadratic Equation

A quadratic equation in $x$ is any equation that can be rearranged into the form $a x^{2} + b x + c = 0$ , where $a$ , $b$ , $c$ are real numbers and $a \neq = 0$ . This layout is called the standard form: every term on the left, zero on the right, and the terms written in falling powers of $x$ . Here $a$ is the coefficient of $x^{2}$ , $b$ is the coefficient of $x$ and $c$ is the constant term. NCERT introduces this in Section 4.2.

The condition $a \neq = 0$ is the heart of the definition. If $a = 0$ , the $x^{2}$ term vanishes and you are left with $b x + c = 0$ , which is linear. But $b$ and $c$ are allowed to be zero. So $2 x^{2} - 18 = 0$ (here $b = 0$ ) and $x^{2} - 7 x = 0$ (here $c = 0$ ) are both quadratic equations.

To recognise a quadratic, never judge by how the equation looks. Expand every bracket, bring all terms to one side and cancel like terms. The equation is quadratic only if the highest power left is exactly 2. For example, $(x + 1)^{3} = x^{3} + 5$ looks cubic, but the $x^{3}$ terms cancel and you get $3 x^{2} + 3 x - 4 = 0$ , a quadratic. On the other hand, $(x + 4)^{2} = x^{2} - 2$ looks quadratic, but the $x^{2}$ terms cancel and you get $8 x + 18 = 0$ , which is linear. If $x$ appears in a denominator, note the values that make it zero, multiply through by the LCM, and then apply the same test.

Why does standard form matter so much? The left side $a x^{2} + b x + c$ is a polynomial of degree 2, so the roots of the equation are exactly the zeroes of that polynomial, and there are at most two of them. Every method in this chapter starts from standard form. Splitting the middle term uses $a c$ and $b$ . The quadratic formula and the discriminant $b^{2} - 4 a c$ use all three coefficients. If you read $b$ as $5$ when it is really $- 5$ , every later step goes wrong.

You may also tidy the equation without changing its roots. Multiplying or dividing both sides by the same non-zero number gives the same roots. So $- x^{2} + 5 x - 6 = 0$ and $x^{2} - 5 x + 6 = 0$ are the same equation. It is usual to make $a$ positive, clear fractions and divide out any common factor. In word problems, you first model the situation with one unknown, then bring the equation to this tidy standard form before solving.

Standard Form of a Quadratic Equation: formulas to remember

Standard form: $a x^{2} + b x + c = 0, a, b, c \in R, a \neq = 0$; Read a, b and c with their signs only after the equation is in this form.
Degree test: $quadratic ⟺ highest power of x after simplifying = 2$; Expand and cancel first; x² or x³ may cancel out.
Condition on a coefficient with k: $(coefficient of x^{2}) \neq = 0$; In questions with an unknown k, the equation is quadratic only for values of k that keep this coefficient non-zero.
Quadratic with no x term (b = 0): $a x^{2} + c = 0$; Still quadratic, because a is not zero.
Quadratic with no constant term (c = 0): $a x^{2} + b x = 0$; Still quadratic; x = 0 is always one of its roots.
Equivalent forms: $a x^{2} + b x + c = 0 ⟺ k (a x^{2} + b x + c) = 0, k \neq = 0$; Multiply by -1 to make a positive, by the LCM to clear fractions, or divide out a common factor.
Condition for a root: $a α^{2} + b α + c = 0$; A number α is a root if substituting it makes the left side zero.
Product of two binomials: $(x + p) (x + q) = x^{2} + (p + q) x + pq$; The fastest way to expand brackets when forming standard form.
Square identities: $(a \pm b)^{2} = a^{2} \pm 2 ab + b^{2}, (a + b) (a - b) = a^{2} - b^{2}$; Used to expand squared brackets correctly; do not write (x - 3)² as x² - 9.
Cube identities: $(a \pm b)^{3} = a^{3} \pm 3 a^{2} b + 3 a b^{2} \pm b^{3}$; Needed when a cubic-looking equation reduces to a quadratic after the x³ terms cancel.

How to solve Standard Form of a Quadratic Equation questions

Use this routine for any question that asks whether an equation is quadratic, asks you to write it in standard form, or asks for a, b and c.

1If x appears in a denominator, write the restrictions first (values of x that make a denominator zero).
2Clear fractions by multiplying both sides by the LCM of the denominators.
3Expand every bracket using the identities, writing each expansion on its own line.
4Bring all terms to the left side so the right side is 0, and collect like terms.
5Check the highest power left: exactly 2 means quadratic; if x² cancels, the equation is linear (or a plain number statement).
6Tidy it: make the x² coefficient positive and divide out any common factor, then write it as ax² + bx + c = 0.
7Read a, b and c with their signs. If the question has an unknown k, state the condition that the x² coefficient is not zero.
8If a value is to be tested, substitute it into the standard form and check that the left side becomes 0; also make sure it is not a restricted value.

Standard Form of a Quadratic Equation 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.

Standard form and identifying a quadratic equation

Suppose your family spends exactly Rs 600 on mangoes at a festival market. The shopkeeper tells you that if the price had been Rs 10 per kg less, you would have got 2 kg more mangoes for the same Rs 600.

Taking the price as Rs x per kg, form an equation for this situation and write it in standard form. State a, b and c. Then decide, by substitution, whether the price was Rs 50 per kg or Rs 60 per kg, and find how many kilograms you bought.

Try it first, then show the solutionSolution

Quantity bought at Rs x per kg $= \frac{600}{x}$ kg.
Quantity at Rs $(x - 10)$ per kg $= \frac{600}{x - 10}$ kg, where $x > 10$ .
Given: $\frac{600}{x - 10} - \frac{600}{x} = 2$ .
Multiply both sides by $x (x - 10)$ : $600 x - 600 (x - 10) = 2 x (x - 10)$ .
Simplify the left side: $6000 = 2 x^{2} - 20 x$ .
Bring all terms to one side: $2 x^{2} - 20 x - 6000 = 0$ .
Divide by 2: $x^{2} - 10 x - 3000 = 0$ , so $a = 1$ , $b = - 10$ , $c = - 3000$ .
Test $x = 50$ : $2500 - 500 - 3000 = - 1000 \neq = 0$ , so Rs 50 is not the price.
Test $x = 60$ : $3600 - 600 - 3000 = 0$ , so Rs 60 per kg fits.
Quantity bought $= \frac{600}{60} = 10$ kg. Check: at Rs 50 per kg you would get $12$ kg, which is 2 kg more.

Answer: Standard form $x^{2} - 10 x - 3000 = 0$ with $a = 1$ , $b = - 10$ , $c = - 3000$ ; the price was Rs 60 per kg and you bought 10 kg.

Memory hook: Fixed money, changing price: two fractions, clear the denominators, and the market sum becomes a quadratic.

Standard form and identifying a quadratic equation

Imagine your class is making a rectangular chart for the science exhibition. The chart's length must be 20 cm more than its width, and the chart must have an area of exactly 1500 square centimetres.

Taking the width as x cm, form the equation in standard form and state a, b and c. Explain why it is a quadratic equation. Then check whether the width is 25 cm or 30 cm, and give the chart's dimensions.

Try it first, then show the solutionSolution

Width $= x$ cm, so length $= (x + 20)$ cm.
Area condition: $x (x + 20) = 1500$ .
Expand: $x^{2} + 20 x = 1500$ .
Standard form: $x^{2} + 20 x - 1500 = 0$ , so $a = 1$ , $b = 20$ , $c = - 1500$ .
It is quadratic because the highest power of $x$ is 2 and $a = 1 \neq = 0$ .
Test $x = 25$ : $625 + 500 - 1500 = - 375 \neq = 0$ , so 25 cm does not work.
Test $x = 30$ : $900 + 600 - 1500 = 0$ , so 30 cm works.
Length $= 30 + 20 = 50$ cm. Check: $30 \times 50 = 1500$ square cm.

Answer: $x^{2} + 20 x - 1500 = 0$ ( $a = 1$ , $b = 20$ , $c = - 1500$ ); the chart is 30 cm wide and 50 cm long.

Memory hook: Length times width equals area: one unknown side, one area, one quadratic.

Tricks to remember Standard Form of a Quadratic Equation

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

Expand, Collect, Check
E-C-C: Expand every bracket, Collect everything on one side, Check the highest power. Only then call it quadratic or not.
Use it when: Any MCQ or assertion-reason item that asks which equation is quadratic.
No a, no quadratic
The letter a is the boss. b and c can be zero and the equation survives, but if a is zero it falls to a linear equation.
Use it when: Questions with k in the x² coefficient, and recognising pure quadratics like 3x² - 12 = 0.
Signs travel with their numbers
Before reading b or c, move everything to the left. Whatever sign a term carries there is the sign of its coefficient.
Use it when: Reading a, b, c from equations like 3x² = 5x - 2, which becomes 3x² - 5x + 2 = 0.
Judge after the dust settles
An x³ on both sides may cancel and leave a quadratic; an x² on both sides may cancel and leave a linear equation. Decide only after cancelling.
Use it when: Cubic-looking or quadratic-looking equations with brackets on both sides.

Standard Form of a Quadratic Equation: solved examples

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

Example 1: Picking the quadratic from four look-alikes · 1 mark · MCQBoard level

Which of the following is a quadratic equation?
(a) $(x - 3)^{2} = x^{2} + 5$
(b) $x^{3} - 2 x^{2} = (x - 1)^{3}$
(c) $(x + 1) (x^{2} + 1) = x^{3} + x^{2} + 4$
(d) $2 x^{2} - x (2 x + 3) = 7$

Try it first, then show the solutionSolution

(a) $x^{2} - 6 x + 9 = x^{2} + 5$ gives $- 6 x + 4 = 0$ : linear.
(b) $(x - 1)^{3} = x^{3} - 3 x^{2} + 3 x - 1$ , so $x^{3} - 2 x^{2} - x^{3} + 3 x^{2} - 3 x + 1 = 0$ , which is $x^{2} - 3 x + 1 = 0$ : quadratic.
(c) Left side $= x^{3} + x^{2} + x + 1$ , so $x + 1 = 4$ , which is $x - 3 = 0$ : linear.
(d) $2 x^{2} - 2 x^{2} - 3 x = 7$ gives $- 3 x - 7 = 0$ : linear.
Only (b) keeps an $x^{2}$ term with nothing higher. [1 mark]

Answer: (b) $x^{3} - 2 x^{2} = (x - 1)^{3}$ , which simplifies to $x^{2} - 3 x + 1 = 0$ .

Example 2: An equation with fractions brought to standard form · 3 marks · Short answerHarder board level

Write the equation $\frac{x}{x + 1} + \frac{x + 1}{x} = \frac{5}{2}$ , where $x \neq = 0, - 1$ , in the standard form $a x^{2} + b x + c = 0$ . State $a$ , $b$ and $c$ , and check whether $x = 1$ is a root.

Try it first, then show the solutionSolution

Restrictions: $x \neq = 0$ and $x \neq = - 1$ , since these make a denominator zero. LCM of denominators $= x (x + 1)$ .
Combine: $\frac{x ^{2} + ( x + 1 ) ^{2}}{x ( x + 1 )} = \frac{5}{2}$ , so $\frac{2 x ^{2} + 2 x + 1}{x ^{2} + x} = \frac{5}{2}$ . [½ mark]
Cross-multiply: $2 (2 x^{2} + 2 x + 1) = 5 (x^{2} + x)$ , that is $4 x^{2} + 4 x + 2 = 5 x^{2} + 5 x$ . [1 mark]
Bring all terms to one side: $x^{2} + x - 2 = 0$ . [½ mark]
So $a = 1$ , $b = 1$ , $c = - 2$ . [½ mark]
Check $x = 1$ : $1 + 1 - 2 = 0$ , and 1 is not a restricted value. In the original: $\frac{1}{2} + \frac{2}{1} = \frac{5}{2}$ . True. [½ mark]

Answer: $x^{2} + x - 2 = 0$ with $a = 1$ , $b = 1$ , $c = - 2$ ; $x = 1$ is a root.

Example 3: When does an equation with k stay quadratic? · 3 marks · Short answerHarder board level

Consider the equation $(k^{2} - 4) x^{2} + (k + 2) x + 3 = 0$ . Find all real values of $k$ for which it is a quadratic equation. For each value of $k$ that is excluded, say what the equation becomes.

Try it first, then show the solutionSolution

For a quadratic, the coefficient of $x^{2}$ must be non-zero: $k^{2} - 4 \neq = 0$ . [1 mark]
$k^{2} \neq = 4$ , so $k \neq = 2$ and $k \neq = - 2$ . [½ mark]
If $k = 2$ : $0 \cdot x^{2} + 4 x + 3 = 0$ , that is $4 x + 3 = 0$ , a linear equation. [1 mark]
If $k = - 2$ : $0 \cdot x^{2} + 0 \cdot x + 3 = 0$ , that is $3 = 0$ , a false statement with no $x$ , so no value of $x$ satisfies it. [½ mark]

Answer: It is quadratic for every real $k$ except $k = 2$ and $k = - 2$ . For $k = 2$ it becomes the linear equation $4 x + 3 = 0$ ; for $k = - 2$ it becomes $3 = 0$ , which has no solution.

Example 4: Proving an equation is always quadratic · 5 marks · Long answerHarder board level

(i) Show that $(x - a) (x - b) + (x - b) (x - c) + (x - c) (x - a) = 0$ is a quadratic equation for all real values of $a$ , $b$ and $c$ , and write it in standard form.
(ii) For $a = 1$ , $b = 1$ , $c = 4$ , write the equation in its simplest standard form and verify that $x = 3$ is a root.

Try it first, then show the solutionSolution

$(x - a) (x - b) = x^{2} - (a + b) x + ab$ . [½ mark]
$(x - b) (x - c) = x^{2} - (b + c) x + b c$ . [½ mark]
$(x - c) (x - a) = x^{2} - (c + a) x + c a$ . [½ mark]
Adding: $3 x^{2} - 2 (a + b + c) x + (ab + b c + c a) = 0$ . [1 mark]
Here the coefficient of $x^{2}$ is $3$ , which is never zero whatever $a$ , $b$ , $c$ are, so the equation is always quadratic. [1 mark]
(ii) $a + b + c = 6$ and $ab + b c + c a = 1 + 4 + 4 = 9$ , so $3 x^{2} - 12 x + 9 = 0$ ; dividing by 3, $x^{2} - 4 x + 3 = 0$ . [1 mark]
Check $x = 3$ : $9 - 12 + 3 = 0$ , so $x = 3$ is a root. [½ mark]

Answer: (i) Standard form $3 x^{2} - 2 (a + b + c) x + (ab + b c + c a) = 0$ , always quadratic since $a = 3 \neq = 0$ . (ii) $x^{2} - 4 x + 3 = 0$ , and $x = 3$ satisfies it.

Example 5: Case study: the school sports ground · 4 marks · Case-basedHOTS / competency

Suppose your school marks out a rectangular sports ground. The PE teacher uses 140 m of boundary tape to go exactly once around it, and the ground covers an area of 1200 square metres. Let the width of the ground be $x$ m.
(i) Express the length of the ground in terms of $x$ . (1 mark)
(ii) Form the equation satisfied by $x$ in standard form and state $a$ , $b$ , $c$ . (1 mark)
(iii) Check whether $x = 30$ and $x = 40$ satisfy the equation. Explain why both can satisfy it although there is only one ground, and give its dimensions. (2 marks)

Try it first, then show the solutionSolution

(i) Perimeter: $2 (length + x) = 140$ , so length $= (70 - x)$ m. [1 mark]
(ii) Area: $x (70 - x) = 1200$ , so $70 x - x^{2} = 1200$ .
Bring all terms to the left and make $a$ positive: $x^{2} - 70 x + 1200 = 0$ , with $a = 1$ , $b = - 70$ , $c = 1200$ . [1 mark]
(iii) $x = 30$ : $900 - 2100 + 1200 = 0$ . True. [½ mark]
$x = 40$ : $1600 - 2800 + 1200 = 0$ . True. [½ mark]
Width 30 m gives length 40 m; width 40 m gives length 30 m. Both describe the same 40 m by 30 m ground with its sides swapped, which is why the quadratic has two roots. [1 mark]

Answer: (i) $(70 - x)$ m (ii) $x^{2} - 70 x + 1200 = 0$ , $a = 1$ , $b = - 70$ , $c = 1200$ (iii) both satisfy it; the ground is 40 m long and 30 m wide.

Where students lose marks in Standard Form of a Quadratic Equation

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

Mistake: Deciding by appearance that an equation is quadratic
The x² terms on both sides can cancel, leaving a linear equation.
Wrong working
$(x + 3)^{2} = x^{2} + 7$ has $x^{2}$ in it, so it is quadratic.
Right working
$x^{2} + 6 x + 9 = x^{2} + 7$ gives $6 x + 2 = 0$ , which is linear, so it is not quadratic.
Do this: Always expand and cancel before checking the highest power.
Usually costs: The full mark in an MCQ or assertion-reason item.
Mistake: Reading b and c before bringing the equation to standard form
Terms on the right change sign when moved to the left.
Wrong working
$3 x^{2} = 5 x - 2$ , so $a = 3$ , $b = 5$ , $c = - 2$ .
Right working
$3 x^{2} - 5 x + 2 = 0$ , so $a = 3$ , $b = - 5$ , $c = 2$ .
Do this: Move everything to the left side first, then read each coefficient with its sign.
Usually costs: Every later step: the factors, formula or discriminant all come out wrong.
Mistake: Thinking an equation with b = 0 or c = 0 is not quadratic
Students confuse 'all three terms present' with the real condition, which is only a ≠ 0.
Wrong working
$2 x^{2} - 8 = 0$ has no $x$ term, so it is not a quadratic equation.
Right working
$2 x^{2} + 0 x - 8 = 0$ with $a = 2 \neq = 0$ , so it is quadratic with $b = 0$ , $c = - 8$ .
Do this: Check only one thing: is the coefficient of x² non-zero after simplifying?
Usually costs: 1 mark in MCQs, and a wrong b in discriminant questions.
Mistake: Forgetting that a k in the x² coefficient must keep it non-zero
The definition's condition a ≠ 0 is skipped when a contains an unknown.
Wrong working
$(k - 1) x^{2} + 3 x + 2 = 0$ is quadratic for every real $k$ .
Right working
It is quadratic only if $k - 1 \neq = 0$ , that is $k \neq = 1$ ; for $k = 1$ it is $3 x + 2 = 0$ , linear.
Do this: Whenever a coefficient of x² has a letter in it, write 'for a quadratic, this coefficient is not zero' as a line.
Usually costs: ½ to 1 mark in 2- and 3-mark questions.
Mistake: Expanding a cube or square bracket wrongly
Students write $(x - 2)^{3}$ as $x^{3} - 8$ or $(x - 3)^{2}$ as $x^{2} - 9$ , dropping the middle terms.
Wrong working
$(x - 2)^{3} = x^{3} - 4$ gives $x^{3} - 8 = x^{3} - 4$ , so $- 8 = - 4$ : no equation.
Right working
$x^{3} - 6 x^{2} + 12 x - 8 = x^{3} - 4$ gives $3 x^{2} - 6 x + 2 = 0$ : quadratic.
Do this: Write the identity first, then substitute, one bracket per line.
Usually costs: The whole question, since the conclusion flips.

Exam tips

In a 'which is quadratic' MCQ, expand every option quickly on rough paper; do not stop at the first option that looks right, because the look-alikes are set deliberately.
When a question with fractions says 'write in standard form', show the restrictions, the LCM line, the cross-multiplication and the final form separately, since step marks are given for each.
Present the final standard form with a positive x² coefficient and no common factor, then list a, b and c clearly; this one line is reused in later sub-parts of case studies.
In questions with k, write the condition on the x² coefficient in words and symbols, and check each excluded value, since the examiner looks for that line.
When asked to verify a root, substitute into the simplified standard form and also into the original equation if it had fractions, and say the value is not a restricted one.

Practice questions on Standard Form of a Quadratic Equation

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

Q1 · 2 marks · Very short answerBoard level
Check whether each of the following is a quadratic equation. Write each in simplified form.
(i) $x (x + 1) + 8 = (x + 2) (x - 2)$
(ii) $(x - 2)^{3} = x^{3} - 4$
Hint
Expand both sides fully and cancel before looking at the highest power. Use the cube identity for (ii).
Answer
(i) $x^{2} + x + 8 = x^{2} - 4$ gives $x + 12 = 0$ : not quadratic (linear). (ii) $x^{3} - 6 x^{2} + 12 x - 8 = x^{3} - 4$ gives $- 6 x^{2} + 12 x - 4 = 0$ , that is $3 x^{2} - 6 x + 2 = 0$ : quadratic. [1 mark each]
Q2 · 3 marks · Short answerHarder board level
Write $\frac{6}{x} - \frac{4}{x + 1} = 1$ , where $x \neq = 0, - 1$ , in standard form. State $a$ , $b$ , $c$ and check whether $x = 3$ and $x = - 2$ are roots.
Hint
Multiply both sides by $x (x + 1)$ . Notice the $x$ terms partly cancel on the left.
Answer
$6 (x + 1) - 4 x = x (x + 1)$ gives $2 x + 6 = x^{2} + x$ , so $x^{2} - x - 6 = 0$ with $a = 1$ , $b = - 1$ , $c = - 6$ . $x = 3$ : $9 - 3 - 6 = 0$ ; $x = - 2$ : $4 + 2 - 6 = 0$ . Neither is a restricted value, so both are roots (check: $2 - 1 = 1$ and $- 3 + 4 = 1$ ).
Q3 · 3 marks · Short answerHarder board level
For what value of $m$ does $(m - 2) x^{3} + m x^{2} + 4 x - 1 = 0$ become a quadratic equation? Write it in standard form for that value. Is there any value of $m$ for which it becomes a linear equation? Give a reason.
Hint
For a quadratic, the $x^{3}$ term must vanish but the $x^{2}$ term must not. For a linear equation, both must vanish at once.
Answer
Need $m - 2 = 0$ , so $m = 2$ ; then the $x^{2}$ coefficient is $2 \neq = 0$ , giving $2 x^{2} + 4 x - 1 = 0$ . For a linear equation we would need $m - 2 = 0$ and $m = 0$ together, which is impossible, so no such value exists.
Q4 · 5 marks · Long answerHarder board level
Represent each situation as a quadratic equation in standard form (do not solve).
(i) A train travels 480 km at a uniform speed. If its speed had been 8 km/h less, it would have taken 2 hours more for the same journey. Take the speed as $x$ km/h.
(ii) The product of two consecutive positive even integers is 288. Take the smaller integer as $x$ .
(iii) Verify that $x = 48$ satisfies your equation in (i), and check it against the situation.
Hint
For (i), time $= \frac{distance}{speed}$ ; the slower time minus the original time is 2 hours.
Answer
(i) $\frac{480}{x - 8} - \frac{480}{x} = 2$ gives $480 x - 480 (x - 8) = 2 x (x - 8)$ , so $3840 = 2 x^{2} - 16 x$ , that is $x^{2} - 8 x - 1920 = 0$ . [2½ marks] (ii) $x (x + 2) = 288$ , so $x^{2} + 2 x - 288 = 0$ . [1½ marks] (iii) $2304 - 384 - 1920 = 0$ . Check: at 48 km/h the time is 10 h; at 40 km/h it is 12 h, which is 2 h more. [1 mark]
Q5 · 2 marks · Very short answerHOTS / competency
Aman says $(x - 1)^{2} + 2 x = x^{2} + 5$ is a quadratic equation because it contains $x^{2}$ . Neha says it is a linear equation. Find who is right, or whether both are wrong, and explain what the equation really is.
Hint
Expand $(x - 1)^{2}$ and watch what cancels: the $x^{2}$ terms and then the $x$ terms.
Answer
$x^{2} - 2 x + 1 + 2 x = x^{2} + 5$ gives $x^{2} + 1 = x^{2} + 5$ , so $1 = 5$ . Both $x^{2}$ and $x$ cancel, leaving a false statement. It is neither quadratic nor linear, and no value of $x$ satisfies it, so both are wrong.

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.What is the standard form of a quadratic equation?Show answer
$a x^{2} + b x + c = 0$ , where $a$ , $b$ , $c$ are real numbers and $a \neq = 0$ .
2.Why must $a \neq = 0$ ?Show answer
If $a = 0$ , the $x^{2}$ term disappears and the equation becomes linear, $b x + c = 0$ .
3.Is $5 x^{2} - 20 = 0$ quadratic? State $b$ .Show answer
Yes, since $a = 5 \neq = 0$ ; here $b = 0$ and $c = - 20$ .
4.Read $a$ , $b$ , $c$ from $4 - x^{2} = 3 x$ .Show answer
Standard form $x^{2} + 3 x - 4 = 0$ , so $a = 1$ , $b = 3$ , $c = - 4$ (or $a = - 1$ , $b = - 3$ , $c = 4$ before tidying).
5.At most how many roots can a quadratic equation have?Show answer
Two, because its left side is a polynomial of degree 2.
6.When is $α$ a root of $a x^{2} + b x + c = 0$ ?Show answer
When $a α^{2} + b α + c = 0$ .

Standard Form of a Quadratic Equation: frequently asked questions

How do I check if an equation is quadratic in Class 10?

Expand all brackets, bring every term to one side and cancel like terms. If the highest power of x left is exactly 2, it is quadratic. Do not decide by looking at the equation before simplifying.

Can b or c be zero in a quadratic equation?

Yes. Only a, the coefficient of x squared, must be non-zero. Equations like 3x² - 12 = 0 or x² - 5x = 0 are quadratic equations.

Does the standard form need a positive coefficient of x squared?

No, any non-zero a is allowed. But multiplying the whole equation by minus 1 does not change its roots, so it is neat and safer to make a positive before solving.

Is an equation with x in the denominator a quadratic equation?

It can become one. After writing the restrictions, multiply through by the LCM and simplify; if the highest power left is 2, it is quadratic. Any root that equals a restricted value must be rejected.

Is completing the square in the 2026-27 syllabus?

No, solving by completing the square is not in the 2026-27 syllabus. You solve quadratic equations by factorisation or by quoting and using the quadratic formula.