NCERT Class 10 Quadratic Equations Solutions | Step-by-Step Exercise Answers Chapter 4

Q: What is the simplest form of a quadratic?

$ax^2 + bx + c = 0$ without common factors and with simplified coefficients.

Q: How can a student quickly check correctness of roots?

Substitute them in the original equation and verify if both sides balance.

Q: What is the best revision strategy for this chapter?

Practice factorization, memorize formulas, and solve multiple word problems to gain confidence.

Q: Why do some equations need rearrangement before solving?

Because methods like factorization or formula application work only in standard form.

Q: What are exam-oriented word problems?

Problems involving numbers, age, geometry, motion, mixtures, and profit that reduce to quadratic equations.

Q: Why is discriminant analysis important in exams?

It quickly determines the nature of roots without solving the full equation.

Q: What is a real-life example of a quadratic equation?

Maximizing area of a rectangular garden using fixed fencing length leads to a quadratic equation.

Q: Does Class 10 require graphing of quadratics?

Only basic understanding; detailed graphing is taught in higher classes.

Q: What is the vertex form of a quadratic?

$y = a(x - h)^2 + k$, where $(h, k)$ is the vertex of the parabola.

Q: Can quadratic equations have irrational roots?

Yes, when $D$ is positive but not a perfect square.

QUADRATIC EQUATIONS-Exercise 4.1

Quadratic Equations in NCERT Class X Mathematics connect algebra to real-world scenarios like area, speed, and geometry problems. Solutions use factorization, completing the square, and quadratic formula, stressing discriminant for root nature and word problem conversion. These step-by-step guides enhance analytical skills, method selection, and CBSE exam readiness for conceptual mastery.

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November 30, 2025 | By Academia Aeternum

QUADRATIC EQUATIONS-Exercise 4.1

Maths - Exercise

Q1. Check whether the following are quadratic equations :

$(x + 1)^2 = 2(x – 3)$
Soultion
$$\begin{aligned}\left( x+1\right) ^{2}&=2\left( x-3\right) \\ x^{2}+1+2x&=2\left( x-3\right) \\ x^{2}+2x+1-2x+6&=0\\ x^{2}+7&=0\\ \Rightarrow x^{2}+0x+7&=0\end{aligned}$$ is a quadratic equation

$x^2 – 2x = (–2) (3 – x)$
Solution:
$$\begin{aligned}x^{2}-2x&=\left( -2\right) \left( 3-x\right) \\ x^{2}-2x&=-6+2x\\ x^2-2x-2x+6&=0\\ x^{2}-4x+6&=0\end{aligned}$$ is a Quadratic equation

$(x – 2)(x + 1) = (x – 1)(x + 3)$
Solution:
$$\begin{aligned}\left( x-2\right) \left( x+1\right) &=\left( x-1\right) \left( x+3\right) \\ x^{2}+x-2x-2&=x^{2}+3x-x-3\\ x^{2}-x^{2}-x-2-2x+3&=0\\ -3x+1&=0\\ 3x-1&=0\end{aligned}$$ is not a quadratic equation

$(x – 3)(2x +1) = x(x + 5)$
Solution:
$$\begin{aligned}\left( x-3\right) \left( 2x-1\right) &=x\left( x+5\right) \\ 2x^{2}+x-6x-3&=x^{2}+5x\\ 2x^{2}-x^{2}-5x-5x-3&=0\\ x^{2}-10x-3&=0\end{aligned}$$ is a quadratic Equation

$(2x – 1)(x – 3) = (x + 5)(x – 1)$
Solution:
$$\begin{aligned}\left( 2x-1\right) \left( x-3\right) &=\left( x+5\right) \left( x-1\right) \\ 2x^{2}-6x-x+3&=x^{2}-x+5x-5\\ 2x^{2}-x^{2}-7x-4x+3+5&=0\\ x^{2}-11x+8&=0\end{aligned}$$ is a quadratic Equation

$x^2 + 3x + 1 = (x – 2)^2$
Solution:
$$\begin{aligned}x^{2}+3x+1&=\left( x-2\right) ^{2}\\ x^{2}+3x+1&=x^{2}-4x+4\\ 3x+4x+1-4&=0\\ 7x-3&=0\end{aligned}$$ is not a quadratic Equation

$(x + 2)^3 = 2x (x^2 – 1)$
Solution:
$$\begin{aligned}\left( x+2\right) ^{3}&=2x\left( x^{2}-1\right) \\ x^{3}+8+3x^{2}\cdot 2+3x\cdot 4&=2x^{3}-2x\\ x^{3}-2x^{3}+8+6x^{2}+12x+4x&=0\\ -x^{3}+6x^{2}+16x+8&=0\end{aligned}$$ is not a quadratic Equation

$x^3 – 4x^2 – x + 1 = (x – 2)^3$
Solution:
$$\begin{aligned}x^{3}-4x^{2}-x+1&=\left( x-2\right) ^{3}\\ x^{3}-4x^{2}-x+1&=x^{3}-8-3\cdot 2x^{2}+3\cdot x\cdot 4\\ -4x^{2}+6x^{2}-x+1-12x+8&=0\\ 2x^{2}-13x+9&=0\end{aligned}$$ is a quadratic Equation

Represent the following situations in the form of quadratic equations :

The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Solution:

Area of a rectangular plot = 528 m²
Let breadth of the plot is $x$
length, which is 1 more than twice of $x$
length = $2x+1$
Area of Rectangular Plot= length $\times$ breadth
$$\begin{aligned}528&=\left( 2x+1\right) x\\ \Rightarrow \left( 2x+1\right) \left( x\right) &=528\\ 2x^{2}+x-528&=0\end{aligned}$$ To factorise by split midde term we need factors of $ac$ such that algebraic sum of factors should be 1 $$2\times 528=33\times 32$$ Also algebraic Sum $$33-32 = 1$$ $$\begin{aligned}2x^{2}+x-528&=0\\ 2x^{2}-32x+33x-52&=0\\ 2x\left( x-16\right) +33\left( x-16\right) &=O\\ \left( x-16\right) \left( 2x+33\right) &=0\\ x-16&=0\\ x&=16\\ 2x+33&=0\\ x&=-\dfrac{33}{2}\end{aligned}$$ therefore breadth of Rectangular Park = 16 m
and length of Park = (16*2)+1= 33 m
The product of two consecutive positive integers is 306. We need to find the integers.
Solution:

Let first number be $x$ next consecutive number = $x+1$
$$\begin{aligned}x\left( x+1\right) &=306\\ x^{2}+x-306&=0\end{aligned}$$
To factorise by split middle term, we need to split 306 into factors so as its algebraic sum is 1
$$306=18\times 17$$ Algebraic sum $$18-17=1$$ $$\begin{aligned}x^{2}+18x-17x-306&=0\\ x\left( x+18\right) -17\left( x+18\right) &=0\\ \left( x+18\right) \left( x-17\right) &=0\\ x-17&=0\\ x&=17\\ x+18&=0\\ x&=-18\end{aligned}$$
Numbers are positive therefore
first number = 17
Second number = 18
Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Solution:

Let Present age of Rohan is $x$
His mother, who is 26 years older, age $x+26$
After 3 years, Rohans age $=x+3$
and his mothers age $=x+29$

Product of their age after 3 years = 360
$$\begin{aligned}\therefore \left( x+3\right) \left( x+29\right) &=360\\ x^{2}+29x+3x+87&=360\\ x^{2}+32x+87-360&=0\\ x^{2}+32x-273&=0\end{aligned}$$
To find factor by split middle term 273 must be factorize to give algebraic sum of 32
Factors of 273 $273=39\times 7$
Algebraic sum $39-7=32$ $$\begin{aligned}\therefore x^{2}+32x-273&=0\\ x^{2}+39x-7x-273&=0\\ x\left( x+39\right) -7\left( x+39\right) &=0\\ \left( x+39\right) \left( x-7\right) &=0\\ x-7&=0\\ x&=7\\ x+39&=0\\ x&=-39\end{aligned}$$
Age can not be negative, therefore Rohan's age = 7 years
A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution:

Let speed of train is $v$
Distance travelled = 480km time taken (t)
$$t=\dfrac{480}{v}$$
If speed of train would have 8km/h less
then time taken
$$\begin{aligned}t+3&=\dfrac{480}{v-8}\\\\ \dfrac{480}{v}+3&=\dfrac{480}{v-8}\\ \dfrac{480+3v}{v}&=\dfrac{480}{v-8}\\ \left( 480+3v\right) \left( v-8\right) &=480v\\ 480v-5440+3v^{2}-24v&=480v\\ -3840+3v^{2}-24v&=0\\ 3v^{2}-24v-3840&=0\\ \Rightarrow v^{2}-8v-1280&=0\end{aligned}$$
To factorize we need factors of 1280 such that their algebraic sum = 8
$$1280=32\times 40$$ algebraic sum $$40-32=8$$ $$\begin{aligned}v^{2}-40v+32v-1280&=0\\ v\left( v-40\right) +32\left( 1v-40\right) &=0\\ \left( v-40\right) \left( v+32\right) &=0\\ v&=40\\ v&=-32\end{aligned}$$
Speed of train 40km/h

Frequently Asked Questions

A quadratic equation is an equation of the form $ax^2 + bx + c = 0$ where $a,\ b\, c$ are real numbers and $a \neq 0$.

If $a = 0$, the equation becomes linear and no longer contains a squared term, so it cannot be quadratic.

The standard form is $ax^2 + bx + c = 0$.

The word “quadratic” comes from “quad,” meaning square, because the highest power of the variable is 2.

The solutions of $ax^2 + bx + c = 0$ are $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

The discriminant $D$ is the expression $b^2 - 4ac$ found inside the square root of the quadratic formula.

It indicates two distinct real roots.

It indicates one real and repeated root.

It indicates no real roots; the solutions are complex.

By splitting the middle term into two terms whose product is (ac), factoring the expression, and using the zero-product property.

If $pq = 0$, then either $p = 0$ or $q = 0$. It is used to solve factored quadratic equations.

It means expressing $bx$ as the sum of two terms whose product equals $ac$, helping in factorization.

It is a method of rewriting a quadratic as a perfect square expression to solve the equation.

It helps derive the quadratic formula and solve equations that are not easy to factor.

Ensure $a = 1$, take half of the coefficient of $x$, square it, add it to both sides, form a perfect square, and solve.

Roots are the values of $x$ that satisfy the equation $ax^2 + bx + c = 0$.

For equation $ax^2 + bx + c = 0$: sum of roots = $-b/a$; product of roots = $c/a$.

The roots are the $x$-intercepts where the parabola $y = ax^2 + bx + c$ crosses the $x$-axis.

When $D = 0$; the parabola is tangent to the x-axis.

When $D < 0$; the graph does not cross or touch the $x$-axis.

It works for all types of quadratic equations, even when factorization is difficult.

They appear in geometry, projectile motion, business profit problems, age problems, and number-based puzzles.

Shape-based problems such as area, diagonal relations, and dimensions often result in a quadratic equation.

They are used in motion under gravity, height-time relations, and projectile trajectories.

By completing the square on the general form $ax^2 + bx + c = 0$.

An expression like $x^2 + 2px + p^2 = (x + p)^2$.

When its discriminant is a perfect square or when integers exist that multiply to $ac$ and sum to $b$.

Using relations involving area, speed, number constraints, or algebraic identities to form $ax^2 + bx + c = 0$.

If length = breadth + 3 and area = 40 sq units, then $b(b+3) = 40$ becomes a quadratic equation.

A quadratic equation without a linear term, i.e., of the form $ax^2 + c = 0$.

Quadratic equations that contain all three terms: $ax^2$, $bx$, and $c$.

A quadratic equation where $a = 1$, e.g., $x^2 + 5x + 6 = 0$.

Substituting answers back ensures the solution satisfies the original problem context.

Factorization-based questions, quadratic formula problems, word problems, discriminant evaluation, and root nature analysis.

Factorization is fastest when applicable; otherwise, the quadratic formula is the safest and most reliable.

Because the discriminant becomes negative, making the square root of a negative number impossible in real numbers.

Forgetting to bring the equation to standard form before applying methods or miscalculating the discriminant.

The sign determines the curve orientation and affects nature of roots.

Rearranging and simplifying the equation so that all terms are on one side of the equal sign.

Yes, when the discriminant is not a perfect square.