What number of exercises are in Chapter 4 for Grade 10?

The fourth chapter of the NCERT Solutions for Class 10 Maths contains four tasks.

Does the quadratic equation matter?

Quadratic functions occupy a special place in the academic curriculum.

Exactly where are quadratic equations used?

Numerous real-world applications of quadratic equations include estimating enclosed space areas, an object's speed, the profit and loss of a product, and curving a piece of equipment for design.

What does a quadratic equation look like in standard form?

A quadratic equation is written in standard form as axe squared + bx plus c is equal to zero.

What does the quadratic formula serve?

One of the top five mathematical formulae is the quadratic formula, which aids in the solution of quadratic equations.

What kind of equation is not quadratic?

When there are more than two solutions to an equation that satisfy it, it is argued that the equation is not quadratic. In essence, the equation is non-quadratic since the square term is missing.

How many solutions may there be for a quadratic equation?

Two roots—or solutions—exist in a quadratic equation with real or complex coefficients. These two options might or might not be separate, and they might or might not be true.

Why have quadratic equations got two solutions?

There are two solutions since a quadratic expression can be expressed as the product of two linear elements, each of which can be equal to zero.

Quadratic Equation Class 10 Chapter 4 Notes

Table of Contents

Quadratic Equation Class 10- Definition

A quadratic equation in the variable x is an equation of ax2+bx+c=0, where a,b, and c are real numbers, a≠0.
For example, 2×2+x−300=0 is a quadratic equation.

Quadratic Equation Class 10- Standard Form

Any equation of the form p(x)=0, where p(x) is a polynomial of degree 2, is a quadratic equation.
But when we write the terms of p(x) in descending order of their degrees, we get the equation’s standard form.
That is, ax2+bx+c=0, a≠0 is called the standard form of a quadratic equation.

Quadratic Equation Class 10- Roots

A solution of the equation p(x)=ax2+bx+c=0, with a≠0, is called a root of the quadratic equation.
A real number α is called a root of the quadratic equation ax2+bx+c=0,a≠0 if aα2+bα+c=0.
It means x=α satisfies the quadratic equation or x=α is the root of the quadratic equation.
The zeroes of the quadratic polynomial ax2+bx+c and the roots of the quadratic equation ax2+bx+c=0 are the same.

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Quadratic Equation Class 10- Method Of Solving

1. Factorisation Method

Factorise the quadratic equation by splitting the middle term.
After splitting the middle term, convert the equation into linear factors by taking common terms out.
Then, on equating each factor to zero, the roots are determined.

For example:

⇒2×2−5x+3 (Split the middle term)

⇒2×2−2x−3x+3 (Take out common terms to determine linear factors)

⇒2x(x−1)−3(x−1)

⇒(x−1)(2x−3) (Equate to zero)\

⇒(x−1)(2x−3)=0

When (x−1)=0 , x=1

When (2x−3)=0 , x=32

So, the roots of 2×2−5x+3 are 1 and 32

2. Method Of Completing The Square

The solution of a quadratic equation can be found by converting any quadratic equation to the perfect square of the form (x+a)2−b2=0.
To convert quadratic equation x2+ax+b=0 to perfect square equate b, i.e., the constant term to the right side of the equal sign, then add a square of half of an, i.e., square of half of the coefficient of x on both sides.
To convert the quadratic equation of form ax2+bx+c=0, a≠0 to perfect square, first divide the equation by an, i.e., the coefficient of x2, then follow the above-mentioned steps.

For example:

⇒x2+4x−5=0 (Equate constant term 5 to the right of the equal sign)

⇒x2+4x=5 (Add a square of half of 4 on both sides)

⇒x2+4x+(42)2=5+(42)2

⇒x2+4x+4=9

⇒(x+2)2=9

⇒(x+2)2−(3)2=0

It is of the form (x+a)2−b2=0

Now,

⇒(x+2)2−(3)2=0

⇒(x+2)2=9

⇒(x+2)=±3

⇒x=1

and x=−5

So, the roots of x2+4x−5=0are 1 and −5

3. By using the quadratic formula

The root of a quadratic equation ax2+bx+c=0 is given by the formula

x=−b±b2−4ac−−−−−−−√2a, where b2−4ac−−−−−−−√ is known as a discriminant.

If b2−4ac−−−−−−−√≥0, then only the root of a quadratic equation is given by

x=−b±b2−4ac−−−−−−−√2a

For example:

⇒x2+4x+3

By using the quadratic formula, we get

⇒x=−4±(4)2−4×1×3−−−−−−−−−−−−−√2×1

⇒x=−4±16−12−−−−−−√2

⇒x=−4±4–√2

⇒x=−4±22

⇒x=−4+22, x=−1

⇒x=−4−22,x=−3

So, the roots of x2+4x+3=0 are −1 and −3

Quadratic Equation Class 10- Nature Of Roots Based On Discriminant

If b2−4ac−−−−−−−√=0, then the roots are real and equal
If b2−4ac−−−−−−−√>0, then the roots are real and distinct.
If b2−4ac−−−−−−−√<0then the roots are imaginary

Learn more about Quadratic Equations

Quadratic Equation Class 10- Sample Questions with Answers

Ques. What will be the nature of the roots of the quadratic equation 2×2 + 4x – n = 0?

Solution: D = b2 – 4ac

⇒ 42 – 4 x 2 (-7)

⇒ 16 + 56 = 72 > 0

Hence, the roots of a quadratic equation are real and unequal.

Ques. If -5 is a root of the quadratic equation 2×2 + px – 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots, then find the value of k.

Solution: Since – 5 is a root of the equation 2×2 + px – 15 = 0

∴ 2(-5)2 + p(-5) – 15 = 0

⇒ 50 – 5p – 15 = 0 or 5p = 35 or p = 7

Again p(x2 + x) + k = 0 or 7×2 + 7x + k = 0 has equal roots

∴ D = 0

i.e., b2 – 4ac = 0 or 49- 4 × 7k = 0

⇒ k = 4928 = 74

Ques. Find the values of k for each of the following quadratic equations so that they have two equal roots.

(i) 2×2 + kx + 3 = 0

(ii) kx (x – 2) + 6 = 0

Solution: (i) We have, 2×2 + kx + 3 = 0

Here, a = 2, b = k, c = 3

D = b2 – 4ac = k2 – 4 × 2 × 3 = k2 – 24 For equal roots

D = 0

i.e., k2 – 24 = 0

⇒ ķ2 = 24

⇒ k = ± √24

⇒ k = + 2√6

(ii) We have, kx(x – 2) + 6 = 0

⇒ kx2 – 2kx + 6 = 0

Here, a = k, b = – 2k, c = 6

For equal roots, we have

D = 0

i.e., b2 – 4ac = 0

⇒ (-2k)2 – 4 × k × 6 = 0

⇒ 4k2 – 24k = 0

⇒ 4k (k – 6) = 0

Either 4k = 0 or k – 6 = 0

⇒ k = 0 or k = 6

But k = 0 6)ecause if k = 0 then given equation will not be a quadratic equation).

So, k = 6.

Ques. If the equation (1 + m2)x2 + 2mcx + c2– a2 = 0 has equal roots, show that c2 = a2 (1 + m2).

Solution: The given equation is (1 + m2) x2 + (2mc) x + (c2– a2) = 0

Here, A = 1 + m2, B = 2mc and C = c2 – a2

Since the given equation has equal roots, therefore D = 0 = B2 – 4AC = 0.

⇒ (2mc)2 – 4(1 + m2) (c2 – a2) = 0

⇒ 4m2c2 – 4(c2 – a2 + m2c2 – m2a2) = 0

⇒ m2c2 – c2 + a2 – m2c2 + m2a2 = 0. [Dividing throughout by 4]

⇒ – c2 + a2 (1 + m2) = 0

⇒ c2 = a(1 + m2) Hence Proved

Ques. If the roots of the quadratic equation (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 are equal, then show that a = b = c.

Solution: Given (x – a) (x – b) + (x – b) (x – c) + (x – 6) (x – a) = 0

⇒ x2 – ax – bx + ab + x2 – bx – cx + bc + x2 – cx – ax + ac = 0

⇒ 3×2 – 2(a + b + c)x + ab + bc + ca = 0

Now, for equal roots D = 0

⇒ B2 – 4AC = 0

⇒ 4(a + b + c)2 – 12(ab + bc + ca) = 0

4a2 + 4b2 + 4c2 + 8ab + 8bc + 8ca – 12ab – 12bc – 12ca = 0

⇒ 2[2a2 + 2b2 + 2co – 2ab – 2bc – 2ca] = 0

⇒ 2[(a2 + b2 – 2ab) + (b2 + c2 – 2bc) + (c2 + a2 – 2ca)] = 0

⇒ [(a – b)2 + (b – c)2 + (c – a)2] = 0

⇒ a – b = 0, b – c = 0, c – a = 0

⇒ a = b, b = c, c = a

⇒ a = b = c

Ques. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in the two subjects.

Solution: Let Shefali’s marks in Mathematics be x.

Therefore, Shefali’s marks in English are (30 – x).

Now, according to the question,

⇒ (x + 2) (30 – x – 3) = 210

⇒ (x + 2) (27 – x) = 210

⇒ 27x – x2 + 54 – 2x = 210

⇒ 25x – x2 + 54 – 210 = 0

⇒ 25x – x2 – 156 = 0

⇒ -(x2 – 25x + 156) = 0

⇒ x2 – 25x + 156 = 0

= x2 – 13x – 12x + 156 = 0

⇒ x(x – 13) – 12(x – 13) = 0