Quadratic Equations in one variable | ICSE Class 10 Maths Notes

This note covers quadratic equations in one variable, standard form, roots, factorisation, the quadratic formula, the discriminant, equal roots, mathematical representation of simple situations, and the interpretation and checking of solutions.

What is a quadratic equation in one variable?

Definition: A quadratic equation in the variable x has the standard form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0.

A variable is a symbol representing a number that can vary or is to be found. Here that symbol is x. The fixed numbers a, b and c are the coefficients: a multiplies x², b multiplies x, and c is the constant term, which contains no variable.

A real number is a number represented on the number line. The notation x² means x multiplied by itself, while ≠ means “is not equal to”. Thus a ≠ 0 requires the coefficient of the squared term to be non-zero.

Why must the squared term remain?

A polynomial is an algebraic expression formed by adding terms whose variable has non-negative integer powers, meaning powers that are whole numbers, zero or greater. Its degree is the highest power with a non-zero coefficient after simplification. A quadratic polynomial has degree two; equating it to zero gives a quadratic equation.

In 2x² − 3x + 1 = 0, the coefficients are a = 2, b = −3 and c = 1. The minus sign belongs to the coefficient b. Reading b as 3 would change the equation and affect every later calculation.

The condition a ≠ 0 preserves degree two. It does not say that b and c must both be non-zero. Identify the squared term and its coefficient after collecting terms, rather than deciding from the appearance of one side alone.

The standard form places the squared term first, the term in x next, and the constant last, with zero on the other side. It provides a consistent way to identify the coefficients before factorising or using a formula.

How do you decide whether an equation is quadratic?

Simplification means expanding brackets and combining like terms, which are terms containing the same power of the variable. Often we need to simplify the given equation before deciding whether it is quadratic. A squared term may disappear when equal terms on opposite sides cancel.

What should you do before reading the coefficients?

1. Expand products and powers on both sides of the equation.
2. Subtract the terms on one side from both sides so that one side becomes zero.
3. Combine terms containing x², terms containing x, and constant terms separately.
4. Check whether the highest remaining power is two with a non-zero coefficient.

Worked example 1. Decide whether (x − 2)² + 1 = 2x − 3 is quadratic.

Answer: Expanding gives x² − 4x + 5 = 2x − 3. Subtracting the right-hand expression gives x² − 6x + 8 = 0. This is quadratic, with a = 1, b = −6 and c = 8.

The expansion of a squared bracket includes its middle term. After expanding, account for the signs of every term moved across the equality. In this example, subtracting −3 adds 3 to the constant already on the left.

Worked example 2. Decide whether x(x + 1) + 8 = (x + 2)(x − 2) is quadratic.

Answer: Expansion gives x² + x + 8 = x² − 4. Subtracting x² from both sides leaves x + 12 = 0. This is a linear equation, an equation of degree one, rather than a quadratic equation.

Can an apparent cubic equation become quadratic?

A cubic equation has degree three after simplification. However, (x + 2)³ = x³ − 4 becomes x³ + 6x² + 12x + 8 = x³ − 4. Cancelling the cubic terms gives 6x² + 12x + 12 = 0.

Dividing by 6 gives x² + 2x + 2 = 0, which is quadratic. Classification depends on the simplified equation, so the largest power seen before expansion does not settle the question.

What are roots, and how do you check them?

Definition: A root of a quadratic equation is a real value of the variable that makes the equation true when substituted. A root is also called a solution of the equation.

Substitution means replacing the variable by a specified value. To test a proposed root, replace every occurrence of x, perform the arithmetic, and compare the two sides. For an equation in standard form, the expression on the left must evaluate to zero.

How are roots related to zeroes?

A zero of a polynomial is a value at which the polynomial evaluates to zero. Consequently, the zeroes of ax² + bx + c and the roots of ax² + bx + c = 0 are the same values.

Result: A quadratic equation has at most two real roots

A quadratic polynomial has at most two zeroes, so its corresponding equation has at most two real roots. “At most” matters: the statement does not promise two different real values. The equation may have equal roots or no real roots.

Worked example 3. Check that 1 is a root of 2x² − 3x + 1 = 0.

Answer: Substituting x = 1 gives 2 × 1² − 3 × 1 + 1 = 2 − 3 + 1 = 0. The equation is satisfied, so 1 is a root.

This check establishes that the proposed value works; it does not establish that all roots have been found. To solve an equation completely, use a method that accounts for both factors or both choices in the quadratic formula.

Keep a clear distinction between the equation, the equality being solved, and its roots, the values that satisfy it. A factorised expression is an intermediate result; the final algebraic answer states the values of x.

How does factorisation give the roots?

Factorisation means writing an expression as a product of factors. A product is the result of multiplication, and the factors are the expressions multiplied together. A linear factor has degree one. If a quadratic polynomial can be written as a product of two linear factors, its equation can be solved by setting each factor equal to zero.

Result: A product equal to zero gives zero factors

When a product of two real expressions is zero, at least one factor is zero. This is the zero-product property. It applies after the equation has been arranged with a product on one side and zero on the other.

How do you split the middle term?

For ax² + bx + c, split bx into two terms whose coefficients add to b and multiply to ac. Then group the terms in pairs and take out common factors. The matching bracket becomes a factor of the whole expression.

Worked example 4. Solve 2x² − 5x + 3 = 0 by factorisation.

Answer: The numbers −2 and −3 add to −5 and multiply to 6, the product of 2 and 3.

1. Write 2x² − 2x − 3x + 3 = 0.
2. Group to obtain 2x(x − 1) − 3(x − 1) = 0.
3. Factorise as (2x − 3)(x − 1) = 0.
4. Set 2x − 3 = 0 or x − 1 = 0, giving x = 3/2 or x = 1.

The common bracket is x − 1. Notice that taking −3 outside the second pair leaves that same bracket. A sign error at this step would produce an expression different from the original polynomial.

Worked example 5. Solve 6x² − x − 2 = 0.

Answer: Split −x as 3x − 4x. Then 6x² + 3x − 4x − 2 = 3x(2x + 1) − 2(2x + 1) = (3x − 2)(2x + 1). Thus 3x − 2 = 0 or 2x + 1 = 0, giving x = 2/3 or x = −1/2.

A negative root is a valid algebraic solution. Restrictions such as positive length come from a word problem, rather than from the factorisation method. Retain both roots until any stated physical conditions have been considered.

How do you use the quadratic formula correctly?

The quadratic formula expresses the roots in terms of the coefficients of the standard form. It is useful when suitable factors are not immediately apparent. First arrange the equation as ax² + bx + c = 0 and identify a, b and c with their signs.

Result: The quadratic formula gives the real roots

Definition: Let D = b² − 4ac, called the discriminant. Provided D ≥ 0, the roots are x = (−b ± √D)/(2a). The symbol ≥ means “greater than or equal to”; √D is the square root of D, meaning the non-negative number whose square is D.

The symbol ± means “plus or minus”. Use it to form two expressions: x = (−b + √D)/(2a) and x = (−b − √D)/(2a). The numerator, the expression above the fraction bar, is divided by the denominator, the expression below it. Here the whole numerator −b ± √D is divided by the denominator 2a.

What is a reliable substitution sequence?

1. Write the coefficients of the standard form, retaining negative signs.
2. Calculate D = b² − 4ac before attempting the square root.
3. If D is non-negative, substitute into (−b ± √D)/(2a).
4. Evaluate the plus and minus branches separately and state both roots.

Worked example 6. Solve 2x² − 5x + 3 = 0 using the formula.

Answer: Here a = 2, b = −5 and c = 3. Therefore D = (−5)² − 4 × 2 × 3 = 25 − 24 = 1. The formula gives x = (5 ± √1)/4 = (5 ± 1)/4. Thus x = 6/4 = 3/2 or x = 4/4 = 1.

The roots agree with factorisation because both methods solve the same equation. In the formula, −b is 5 when b is −5, while b² is 25. These are separate operations, so write them separately before simplifying.

Worked example 7. Solve 2x² − 6x + 3 = 0, leaving the roots in exact form.

Answer: With a = 2, b = −6 and c = 3, D = 36 − 24 = 12. Therefore x = (6 ± √12)/4 = (6 ± 2√3)/4 = (3 ± √3)/2.

Exact form retains the square root without rounding. The two answers here are (3 + √3)/2 and (3 − √3)/2. Keeping the fraction brackets makes the scope of the division clear throughout the calculation.

How does the discriminant determine the nature of roots?

The nature of the roots describes whether there are two distinct real roots, two equal real roots, or no real roots. “Distinct” means different; “equal” means the two roots have the same value. Calculate D from the coefficients of the simplified standard form.

Result: The sign of the discriminant classifies the roots

For a positive discriminant, the plus and minus branches differ because √D is non-zero and a is non-zero. For a zero discriminant, the branches coincide. These equal roots are also called coincident roots or a repeated root.

Worked example 8. Find the discriminant and nature of the roots of 2x² − 4x + 3 = 0.

Answer: Here a = 2, b = −4 and c = 3. Then D = (−4)² − 4 × 2 × 3 = 16 − 24 = −8. Since D < 0, the equation has no real roots.

Keep the word real in this conclusion. The classification concerns real roots, and the square root of a negative discriminant cannot be evaluated as a real number. There is no need to continue substituting into the real-root formula.

Does a negative discriminant change the degree?

No. The equation in the example remains quadratic because its squared coefficient is non-zero. Its degree describes its algebraic form; the discriminant describes its roots. A quadratic equation does not stop being quadratic because no real value satisfies it.

If a question asks only for the nature of roots, the coefficients, discriminant calculation and classification supply the required reasoning. If it also asks for the roots when real, continue with factorisation or the formula after classifying them.

How do equal roots help you find an unknown coefficient?

Equal roots occur when D = 0. This gives an equation involving the coefficients. If a coefficient contains an unknown constant, use the zero-discriminant condition to find its possible values, while keeping the leading coefficient, the coefficient of x², non-zero.

Worked example 9. Find the discriminant and roots of 3x² − 2x + 1/3 = 0.

Answer: Here a = 3, b = −2 and c = 1/3. Hence D = (−2)² − 4 × 3 × (1/3) = 4 − 4 = 0. Both roots are −b/(2a) = 2/6 = 1/3.

Writing the roots as 1/3 and 1/3 records the repetition. There is one distinct real value, but the conventional description is “two equal real roots”. These statements describe the same situation and should not be confused.

What changes when a coefficient is a parameter?

A parameter is a fixed but unspecified number in an equation. In the following example, k is the parameter and x remains the variable being solved for. The task is to determine k from the requirement that the roots be equal.

Worked example 10. Find the values of k for which 2x² + kx + 3 = 0 has two equal roots.

Answer: The coefficients are a = 2, b = k and c = 3. Therefore D = k² − 4 × 2 × 3 = k² − 24. Equal roots require k² − 24 = 0, so k² = 24 and k = ±2√6.

The plus and minus signs here give two possible parameter values. They are not the roots x of one equation. Each choice fixes a different equation whose two roots are equal. The leading coefficient is 2 in both cases, so both equations remain quadratic.

How do you form and solve a quadratic equation for dimensions?

In a mathematical model, an equation expresses the relationships given in a situation. Choose the unknown quantity, state its units, express related quantities using the same variable, and translate the stated relationship into an equation.

How does a rectangular area lead to a quadratic?

Consider a rectangular prayer hall of area 300 square metres. Its length is one metre more than twice its breadth. The breadth is the shorter side in this situation. Let x be the numerical value of the breadth in metres.

The length is then (2x + 1) metres. The abbreviation m means metre, and m² means square metre, a unit of area. Multiplying the length by the breadth gives x(2x + 1) = 300, or 2x² + x − 300 = 0.

The rectangle has x beside a vertical side, 2x + 1 below a horizontal side, and 300 m² inside. The labels connect the two dimensions with the given area.

See Fig. 4.1 in your NCERT textbook

Worked example 11. Find the dimensions of a rectangular hall with area 300 m² and length one metre more than twice its breadth.

Answer: Taking its breadth as x metres gives 2x² + x − 300 = 0.

1. Split the middle term: 2x² − 24x + 25x − 300 = 0.
2. Group the terms: 2x(x − 12) + 25(x − 12) = 0.
3. Factorise: (x − 12)(2x + 25) = 0, so x = 12 or x = −12.5.
4. The breadth must be positive, so take x = 12. The length is 2 × 12 + 1 = 25 metres.

Why must you interpret the solutions?

The negative value satisfies the algebraic equation but cannot represent the breadth of the hall. Reject it for that stated reason. The final answer is breadth 12 m and length 25 m, identifying both quantities with their units.

Check both pieces of original information: 12 × 25 = 300 gives the area, and 25 = 2 × 12 + 1 gives the length relationship. Checking only the quadratic equation would leave the physical interpretation unexamined.

How do you translate number and production problems into equations?

Not every quadratic model concerns lengths. A product is the result of multiplication, and a stated product can connect two expressions involving the same unknown. Carefully distinguish an original quantity from its value after a stated change.

How do changes in two quantities affect their product?

John and Jivanti initially have 45 marbles altogether. Each loses 5 marbles, and the product of their remaining numbers is 124. Let x be John's original number. Jivanti's original number is 45 − x because their total is fixed.

After the losses, John has x − 5 and Jivanti has 40 − x. The product condition applies to these remaining numbers. It therefore gives (x − 5)(40 − x) = 124, rather than an equation using their original numbers.

Worked example 12. Form the quadratic equation for the marble situation: the original total is 45, each person loses 5, and the product of the remaining numbers is 124. Let x be John's original number.

Answer: (x − 5)(40 − x) = 124 expands to −x² + 45x − 200 = 124. Bringing the constant across gives −x² + 45x − 324 = 0, or x² − 45x + 324 = 0.

How do quantity and cost combine?

A cottage industry produces toys. If x is the number produced in a day, the stated cost of producing each toy is 55 − x rupees. The total production cost equals the number produced multiplied by the cost of each toy.

For a total cost of 750 rupees, write x(55 − x) = 750. Expanding gives 55x − x² = 750, so the standard form is x² − 55x + 750 = 0. The unit rupee measures money, and its symbol is ₹.

In both examples, defining x precisely determines the meaning of every later expression. Forming the equation answers a representation question. If the question asks for the quantities themselves, solving and interpreting the resulting roots are further necessary steps.

How can a quadratic equation test whether a geometric arrangement is possible?

A quadratic equation can connect a geometric condition with the existence of suitable lengths. Consider a circular park of diameter 13 m, with gates A and B at opposite ends of that diameter. A pole is to stand on the boundary with its distances from the gates differing by 7 m.

How does the diagram provide an equation?

A diameter is a straight segment through a circle's centre with endpoints on the circle. Let P label the pole's position. The notation AP means the length of the segment from A to P; BP and AB are interpreted in the same way.

A, B and P lie on the circle. AB is the diagonal diameter labelled 13; AP and PB complete the triangle.

See Fig. 4.2 in your NCERT textbook

The angle at P is a right angle, an angle of 90 degrees, because AB is a diameter. The hypotenuse is the side opposite a right angle, here AB. Pythagoras' theorem states that its square equals the sum of the squares of the other two sides.

Choose BP as the shorter distance and let it be x metres. Then AP is (x + 7) metres. The relation AP² + BP² = AB² gives (x + 7)² + x² = 13², which simplifies to x² + 7x − 60 = 0.

Worked example 13. For the circular park with diameter 13 m and difference of pole-to-gate distances 7 m, determine suitable distances.

Answer: The shorter distance x satisfies x² + 7x − 60 = 0. Here D = 7² − 4 × 1 × (−60) = 289, which is positive. The formula gives x = (−7 ± √289)/2 = (−7 ± 17)/2, so x = 5 or x = −12. Reject the negative distance. The distances are 5 m and 12 m.

What does this example show about feasibility?

Feasibility means whether the stated arrangement can be achieved. The positive discriminant supplies real algebraic possibilities here, and the positive root supplies the required lengths. Their difference is 7 m, and their squared lengths add to the square of the diameter.

The answer depends on both algebra and interpretation. In a problem about measurements, finding real roots is followed by checking their suitability as measurements. State why an inadmissible root is discarded and return to the quantities named in the question.

Glossary

• Quadratic equation — An equation that simplifies to ax² + bx + c = 0, with real coefficients and a non-zero.
• Standard form — The arrangement ax² + bx + c = 0, with terms in descending powers and a non-zero.
• Coefficient — A fixed number multiplying a variable term, including the sign attached to that number.
• Degree — The highest power of the variable with a non-zero coefficient after simplifying a polynomial.
• Root — A value of the variable that makes the equation true when substituted into it.
• Zero of a polynomial — A value of the variable for which the polynomial evaluates to zero.
• Factorisation — Rewriting an algebraic expression as a product of factors without changing its value.
• Linear factor — A factor that is a polynomial of degree one in the variable.
• Quadratic formula — The expression (−b ± √(b² − 4ac))/(2a), giving real roots when the discriminant is non-negative.
• Discriminant — The quantity b² − 4ac, whose sign determines the nature of a quadratic equation's roots.
• Distinct roots — Two different real values satisfying a quadratic equation with positive discriminant.
• Equal roots — Two coincident real roots occurring when the discriminant is zero, both equal to −b/(2a).
• Parameter — A fixed but unspecified number in an equation, sometimes determined by a condition on its roots.
• Mathematical model — An equation or set of expressions representing the relationships given in a situation.

Common errors and misconceptions

• Misconception: Seeing x² on either side proves that an equation is quadratic. Correct: Expand and simplify first; squared terms can cancel, leaving a linear equation.
• Misconception: The coefficient of x in 2x² − 5x + 3 = 0 is 5. Correct: It is −5. Retain that sign when calculating b² and −b.
• Misconception: Once the expression is factorised, solving is complete. Correct: Set each factor equal to zero and solve the resulting linear equations.
• Misconception: In the formula, only the square-root term is divided by 2a. Correct: The whole numerator −b ± √D is divided by 2a.
• Misconception: Every quadratic has two different real roots. Correct: The discriminant distinguishes two distinct real roots, two equal real roots and no real roots.
• Misconception: D = 0 means there are no real roots. Correct: It gives two equal real roots, each −b/(2a); no real roots occur when D is negative.
• Misconception: A negative root can be discarded in every problem. Correct: Negative roots are algebraically valid; discard a root when it contradicts a stated contextual requirement, such as positive breadth.
• Misconception: The product after two people lose marbles can use their original numbers. Correct: Subtract each person's loss before forming the product equation.

Exam-style questions with model answers

Q1. Is x(x + 1) + 8 = (x + 2)(x − 2) a quadratic equation? Justify your answer by simplifying it. [2 marks]

1. Expanding both sides gives x² + x + 8 = x² − 4.
2. Cancelling x² and rearranging gives x + 12 = 0. Its degree is one, so it is linear, not quadratic.

Q2. Solve 2x² − 5x + 3 = 0 by factorisation, showing the middle-term split. [3 marks]

1. The coefficients −2 and −3 add to −5 and multiply to 6. Split the middle term to obtain 2x² − 2x − 3x + 3 = 0.
2. Grouping gives 2x(x − 1) − 3(x − 1) = 0, so the equation becomes (2x − 3)(x − 1) = 0.
3. At least one factor is zero. Hence 2x − 3 = 0 or x − 1 = 0, giving the roots x = 3/2 and x = 1.

Q3. Use the discriminant to determine the nature of the roots of 2x² − 4x + 3 = 0. [3 marks]

1. Comparing with the standard form ax² + bx + c = 0 gives a = 2, b = −4 and c = 3. The negative sign is part of b.
2. The discriminant is D = b² − 4ac = (−4)² − 4 × 2 × 3 = 16 − 24 = −8.
3. Since D is negative, the equation has no real roots. No real number has a square equal to −8, so the real-root formula gives no real solution.

Q4. Find all values of the parameter k for which 2x² + kx + 3 = 0 has two equal real roots. [3 marks]

1. In standard form, a = 2, b = k and c = 3. Equal real roots require the discriminant b² − 4ac to be zero.
2. Therefore k² − 4 × 2 × 3 = 0, which simplifies to k² − 24 = 0 and hence k² = 24.
3. Taking both square-root choices gives k = ±√24 = ±2√6. Both values are admissible because the leading coefficient remains 2, so each resulting equation is quadratic.

Q5. Solve 2x² − 6x + 3 = 0 using the quadratic formula. Give exact roots and state their nature. [4 marks]

1. The equation is already in standard form, with a = 2, b = −6 and c = 3. Thus the denominator 2a in the formula is 4.
2. Calculate the discriminant: D = (−6)² − 4 × 2 × 3 = 36 − 24 = 12. Because D is positive, there are two distinct real roots.
3. Substitute into the formula to obtain x = (6 ± √12)/4. Since √12 = 2√3, this becomes x = (6 ± 2√3)/4.
4. Divide numerator and denominator by 2. The exact roots are x = (3 + √3)/2 and x = (3 − √3)/2.

Q6. A rectangular prayer hall has area 300 m². Its length is one metre more than twice its breadth. Form a quadratic equation and find both dimensions, explaining any rejected root. [5 marks]

1. Let x be the breadth in metres. Since the length is one metre more than twice the breadth, its length is (2x + 1) metres.
2. The rectangular area gives x(2x + 1) = 300. Expanding and rearranging yields the quadratic equation 2x² + x − 300 = 0.
3. Split the middle term: 2x² − 24x + 25x − 300 = 0. Grouping gives 2x(x − 12) + 25(x − 12) = 0.
4. Thus (x − 12)(2x + 25) = 0, giving x = 12 or x = −12.5. Reject −12.5 because the breadth must be positive.
5. The breadth is 12 m and the length is 2 × 12 + 1 = 25 m. Their product is 300 m², confirming the required area.

Q7. Gates A and B are at opposite ends of a 13 m diameter of a circular park. A pole P is placed elsewhere on the boundary with AP longer than BP by 7 m. Use the right angle at P to form a quadratic equation, then find AP and BP using the quadratic formula. [5 marks]

1. Let BP = x metres, so AP = (x + 7) metres. Since the triangle is right-angled at P, Pythagoras' theorem gives x² + (x + 7)² = 13².
2. Expanding gives 2x² + 14x + 49 = 169. Rearranging and dividing by 2 gives x² + 7x − 60 = 0.
3. The discriminant is D = 7² − 4 × 1 × (−60) = 49 + 240 = 289, which is positive, so the equation has two real roots.
4. The quadratic formula gives x = (−7 ± √289)/2 = (−7 ± 17)/2. Therefore the algebraic roots are x = 5 and x = −12.
5. Reject −12 because a distance must be positive. Hence BP = 5 m and AP = 12 m, whose difference is the required 7 m.

Q8. Find the discriminant of 3x² − 2x + 1/3 = 0, state the nature of its roots, and find them. [3 marks]

1. The coefficients are a = 3, b = −2 and c = 1/3. Therefore D = b² − 4ac = (−2)² − 4 × 3 × (1/3) = 4 − 4 = 0.
2. A zero discriminant gives two equal real roots. Both branches of the quadratic formula coincide because adding or subtracting the square root of zero makes no difference.
3. Each root equals −b/(2a) = 2/(2 × 3) = 1/3. Thus the roots are 1/3 and 1/3, with one distinct real value.

Key takeaways

• A quadratic equation simplifies to ax² + bx + c = 0, where the coefficients are real and a is non-zero.
• Expand and collect terms before classifying an equation, because squared terms or cubic terms can cancel.
• A root makes the equation true on substitution; roots of the equation are zeroes of its corresponding polynomial.
• Factorisation solves a quadratic by writing a product equal to zero and setting each linear factor equal to zero.
• The quadratic formula gives both real roots, provided the discriminant is non-negative; the entire numerator is divided by 2a.
• A positive, zero or negative discriminant gives distinct real roots, equal real roots or no real roots respectively.
• Define the unknown quantity before forming a word-problem equation, and express other quantities consistently using that variable.
• Interpret each algebraic solution in its original context, reject unsuitable values with reasons, and state measurement answers with units.

Test yourself

Why is a ≠ 0 required in ax² + bx + c = 0?

The squared term must have a non-zero coefficient for the simplified equation to have degree two.

What are a, b and c in 2x² − 3x + 1 = 0?

They are a = 2, b = −3 and c = 1; the minus sign belongs to b.

What does substituting x = 1 into 2x² − 3x + 1 = 0 show?

The left side becomes 2 − 3 + 1 = 0, proving that 1 is a root.

What are the roots if (3x − 2)(2x + 1) = 0?

Setting each factor to zero gives x = 2/3 or x = −1/2.

What do D = 0 and D < 0 mean for a quadratic equation?

D = 0 gives two equal real roots; D < 0 gives no real roots.

Which condition finds k when 2x² + kx + 3 = 0 has equal roots?

Set the discriminant to zero: k² − 24 = 0, giving k = ±2√6.

Why is −12.5 rejected when the hall's breadth satisfies 2x² + x − 300 = 0?

Although it is an algebraic root, a negative number cannot represent the hall's breadth in metres.

John and Jivanti start with 45 marbles altogether and each loses 5. If John starts with x, what remains with each?

John has x − 5 marbles, and Jivanti has 45 − x − 5 = 40 − x marbles.