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

⇒ (x – 13) (x – 12) = 0

Either x – 13 or x – 12 = 0

∴ x = 13 or x = 12

Therefore, Shefali’s marks in Mathematics = 13

Marks in English = 30 – 13 = 17

or Shefali’s marks in Mathematics = 12

marks in English = 30 – 12 = 18.

Ques. The sum of the areas of two squares is 468 m2. If the difference in their perimeters is 24 m, find the sides of the two squares.

Solution: Let x be the length of the side of the first square and y be the length of the side of the second square.

Then, x2 + y2 = 468 …(i)

Let x be the length of the side of the bigger square.

4x – 4y = 24

⇒ x – y = 6 or x = y + 6 …(ii)

Putting the value of x in terms of y from equation (ii), in equation (i), we get

(y + 6)2 + y2 = 468

⇒ y2 + 12y + 36 + y2 = 468 or 232 + 12y – 432 = 0

⇒ y2 + 6y – 216 = 0

⇒ y2 + 18y – 12y – 216 = 0

⇒ y(y + 18) – 12(y + 18) = 0

⇒ (y + 18)(y – 12) = 0

Either y + 18 = 0 or y – 12 = 0

⇒ y = -18 or y = 12

But, sides cannot be negative, so y = 12

Therefore, x = 12 + 6 = 18

Hence, the sides of the two squares are 18 m and 12 m.