Quadratic Equation Class 10th

Definition

The polynomial in one variable (like x) with degree 2 is called Quadratic Equation.

Examples – 2x² + 3x + 5, x² – 4x + 7, x² + 8, 3x² + 9x,…. etc.

Standard Form of Quadratic Equation

ax² + bx + c = 0

Where a, b, and c are real numbers and a ≠ 0

x = Variable (unknown)

Some Important Points

• In the quadratic equation, the highest power of variable x is 2 so it is also called the quadratic polynomial.
• The degree of any polynomial shows the number of solutions which are called zeroes or roots.
• In the quadratic equation, the degree is 2 so roots (solutions) will be always two values.
• In any quadratic equation, coefficients b and c can be zero but a can’t be zero because if a = 0 then that polynomial will not be quadratic that will be linear.

Examples –

2x2 + 3x + 5	a = 2, b = 3, c = 5
x2 – 4x + 7	a = 1(we generally do not write 1x2), b = -4, c = 7
x2 + 8	a = 1, b = 0, c = 8
3x2 + 9x	a = 3, b = 9, c = 0

Solution of Quadratic Equation

Factorization Method

In this method, we factorize the quadratic equation and put each factor equal to zero and then find the values of x. These values of x are the solution of the quadratic equation and are called the roots of the Quadratic Equation.

This can be understood by the following examples.

Example – 1) Find the roots of the quadratic equation x² + 5x + 6 = 0 by factorization method.

Solution – given equation   x² + 5x + 6 = 0

Step (1) Comparing the equation with the standard form ax² + bx + c = 0

a = 1, b = 5, c = 6

We take      a⨯c = 1⨯6 = 6    and      b = 5

Step (2) Now we have to take two numeric values so that their sum should be equal to b i.e., 5 and their multiplication should be equal to a⨯c i.e., 6. So, we take the numbers 2 and 3.

Note – If it is difficult to find the two numeric values, we can factorize the term a⨯c and make pair to get it easily.

Step (3) now check

2+3 = 5 and 2⨯3 = 6 (these values satisfy)

Step (4) Now break the middle term of equation x2 + 5x + 6 with the help of the above sum 2+3 = 5

x² + (2+3)x + 6 = 0

x² + 2x + 3x + 6 = 0

Step (5) Now take common terms out from the first two terms and last two terms.

x(x + 2) + 3(x+2) = 0

(x+2) is again a common term so

(x+2) (x+3) = 0

Step (6) Now, by putting each factor equal to zero.

(x+2) = 0   and    (x+3) = 0

x = −2    and         x = −3 Ans.

Both the values of x are the solution of the given quadratic equation and are called the roots of this equation.

We can check our answer by putting the values of x in the given quadratic equation.

x² + 5x + 6 = 0

at x = −2,                                                          

(−2)² + 5(−2) + 6 = 0                                         

4 – 10 + 6 = 0                                                  

– 6 + 6 = 0                                                         

0 = 0                                                                 

LHS = RHS                                                    

at x = −3,

(−3)² + 5(−3) + 6 = 0

9 – 15 + 6 = 0

– 6 + 6 = 0

0 = 0

LHS = RHS

At both values, LHS = RHS which means our answer is correct.

Example – 2) Find the roots of the quadratic equation 2x² – 5x + 3 = 0 by factorization method.

Solution – 2x² – 5x + 3 = 0              Here, a⨯c = 2⨯3 = 6  and  b = −5

2x² – (2+3)x + 3 = 0                         two numeric values −2 and −3

2x² – 2x − 3x + 3 = 0                       −2⨯(−3) = 6   and  −2 + (-3) = −2 − 3 = −5 

2x(x − 1) − 3(x − 1) = 0

(x − 1) (2x − 3) = 0

(x − 1) = 0   and   (2x − 3) = 0

x = 1    and       x = 3/2     

Thus, the roots of the quadratic equation 2x² – 5x + 3 = 0 are x = 1 and x = 3/2.    Ans.

Perfect The Square Method (Complete The Square Method)

In this method, we convert the quadratic equation into the perfect square form and then we take the square root to obtain the required roots.

This method will be clear by the following examples.

Example – 1) Solve the quadratic equation 9x² – 3x – 2 = 0 by the perfect square method.

Solution – Given equation 9x² – 3x – 2 = 0

Step 1) We divide the whole equation by 9 to make 1 as the coefficient of x².  

9x²/9 – 3x/9 – 2/9 = 0

x² – x/3 – 2/9 = 0

Step 2) Taking the constant term to the right side,

x² – x/3 = 2/9  

Step 3) Adding the square of half of the coefficient of x to both sides to make a perfect square on the left side.

x² – x/3 + (⅙)² = 2/9 + (⅙)²             [Half of coefficient of x = ⅓ ÷ 2 = ⅓⨯½ = ⅙]

Step 4) Now the left side is a perfect square

(x – ⅙)² = 2/9 + 1/36

(x – ⅙)² = (2⨯4 + 1⨯1)/36 (LCM = 36)

(x – ⅙)² = 9/36

(x – ⅙)² = ¼

Easy Tip – To make a perfect square, we put the sign of the middle term (– x/3) of the equation (x² – x/3 – 2/9 = 0) which in this example is – (negative), between the variable x and half of the coefficient of x which is ⅙, and make a perfect square (x – ⅙)².

Step 5) By taking the square root of both sides.

√(x – ⅙)² = √¼    

x – ⅙ = ±½

Taking both the values separately,

On taking the (+) sign,

x – ⅙ = +½                                            

x = ½ + ⅙                                                          

x = (1⨯3 + 1⨯1)/6                                               

x = 4/6                                               

x = ⅔                                             

On taking the (−) sign,

x – ⅙ = −½

x = −½ + ⅙ 

x = (−1⨯3 + 1⨯1)/6 

x = −2/6

x = −⅓

So, x = ⅔ and x = −⅓ are the required roots of the quadratic equation 9x² – 3x – 2 = 0.    Ans.

Example – 2) Solve the quadratic equation 2x² – 7x + 3 = 0 by the perfect square method.

Solution – Given equation 2x² – 7x + 3 = 0

Dividing the equation by 2 to make 1 as the coefficient of x²

2x²/2 – 7x/2 + 3/2 = 0

x² – 7x/2 + 3/2 = 0

Taking the constant term to the right side

x² – 7x/2 = –3/2 

Adding the square of half of the coefficient of x to both side   

x² – 7x/2 + (7/4)² = –3/2 + (7/4)² [Half of coefficient of x = 7/2 ÷ 2 = 7/2⨯½ = 7/4]

(x – 7/4)² = –3/2 + 49/16 = (–3⨯8 + 49⨯1)/16 = (–24 + 49)/16 = 25/16

(x – 7/4)² = 25/16  

Taking square root on both sides,

√(x – 7/4)² = √(25/16)    

x – 7/4 = ±5/4

On taking the (+) sign,

x – 7/4 = +5/4                       

x = 5/4 + 7/4                          

x = (5 + 7)/4 = 12/4                          

x = 3

On taking the (−) sign, 

x – 7/4 = –5/4

x = –5/4 + 7/4

x = (–5 + 7)/4 = 2/4

x = ½

Both the values of the x are the required roots.  Ans.

Quadratic Formula (Shridharacharya Sutra)

The quadratic formula is given by the great Indian mathematician Shridharacharya and is also known as Shridharacharya Sutra. This formula is used to solve the quadratic equation.

x = –b ± √(b² – 4ac)/2a

Where x = variable

a, b, c = Coefficients of quadratic equation

↣ This formula can be derived from the perfect square method.

The standard form of a quadratic equation is

ax² + bx + c = 0

x² + bx/a + c/a = 0  (taking 1 as coefficient of x² or dividing the whole equation by the coefficient of x² (a))

x² + bx/a = −c/a       (taking the constant term to the right side)

x² + bx/a + (b/2a)² = −c/a + (b/2a)²   (adding square of half of coefficient of x)

(x + b/2a)² = −c/a + b²/4a²  

(x + b/2a)² = (−c⨯4a + b²⨯1)/4a² = (−4ac + b²)/4a²

Taking square root on both sides,

√(x + b/2a)² = √(−4ac + b²)/4a²    

x + b/2a = ±√(b² − 4ac)/2a   

x = −b/2a ± √(b² − 4ac)/2a

x = −b ± √(b² − 4ac)/2a

Here, sign ± shows the two values of x.

x = −b + √(b² − 4ac)/2a                           x = −b − √(b² − 4ac)/2a

The above values of x are the Shridharacharya Sutra.

Example – Solve the quadratic equation 9x² + 7x – 2 = 0 by Shridharacharya sutra.

Solution – given equation 9x² + 7x – 2 = 0

Comparing with standard form ax² + bx + c = 0

So, a = 9, b = 7, c = – 2

By quadratic formula,  x = −b ± √(b² − 4ac)/2a 

Putting the values of a, b, c

x = −7 ± √{7² − 4⨯9⨯(-2)}/2⨯9

x = −7 ± √{49 + 72}/18

x = −7 ± √{121}/18 = −7 ± 11/18  

On taking the (+) sign,

x = −7 + 11/18                              

x = 4/18                                      

x = 2/9                                           

On taking the (−) sign,

x = −7 − 11/18

x = −18/18

x = −1            

These values of x are the roots of the given quadratic equation.    Ans.

Discriminant and Nature of Roots

We can find the nature of roots by the term b² – 4ac used in the Shridharacharya sutra which is called the discriminant. It is denoted by D.

D = b² – 4ac

⇒ If D > 0 then the roots will be different and real.

x = (−b + √D)/2a       and       x = (−b − √D)/2a

⇒ If D = 0 then the roots will be equal and real.

x = −b/2a       and       x = −b/2a

⇒ If D < 0 then the roots will not be real, they will be imaginary.

Example – 1) find the nature of the roots of equation 2x² – 6x + 3 = 0 and if roots exist then find them.

Solution – Given equation 2x² – 6x + 3 = 0

Comparing the equation with ax² + bx + c = 0

Here, a = 2, b = – 6, c = 3

Now D = b² – 4ac

D = (-6)² – 4⨯2⨯3 = 36 – 24 = 12

∵ D > 0

∴ The roots of equation 2x² – 6x + 3 = 0 will be different and Real.

By Shridharacharya formula,  x = (−b ± √D)/2a        where, D = b² – 4ac

x = {−(-6) ± √12)}/2⨯2 = {6 ± √(4⨯3)}/4

x = {6 ± 2√3}/4 = 2{3 ± √3}/4 = {3 ± √3}/2                     

So, the roots are x = (3 + √3)/2 and x = (3 − √3)/2.       Ans.

Example – 2) find the nature of roots of the equation x² – 4x + 4 = 0 and if roots exist then find them.

Solution – Given equation x² – 4x + 4 = 0

By comparing with ax² + bx + c = 0

Here, a = 1, b = – 4, c = 4

Now D = b² – 4ac

D = (-4)² – 4⨯1⨯4 = 16 – 16 = 0

∵ D = 0

∴ The roots of the equation x² – 4x + 4 = 0 will be equal and Real.

By quadratic formula,     x = (−b ± √D)/2a        where, D = b² – 4ac

x = {−(-4) ± √0)/2⨯1 = (4 ± 0)/2 = 4/2

x = 2

So, the roots are x = 2 and x = 2.       Ans.

Example – 3) find the nature of the roots of equation 2x² + 3x + 5 = 0 and if roots exist then find them.

Solution – Given equation 2x² + 3x + 5 = 0

On comparing with ax² + bx + c = 0

a = 2, b = 3, c = 5

Now D = b² – 4ac

D = (3)² – 4⨯2⨯5 = 9 – 40 = –31

∵ D < 0

∴ The roots of equation 2x² + 3x + 5 = 0 will not be Real. They will be imaginary.

By Shridharacharya quadratic formula,   x = (−b ± √D)/2a       where, D = b² – 4ac

x = (−3 ± √-31)/2⨯2 = (−3 ± √−1⨯31)/4       

x = (−3 ± √−1⨯√31)/4 = (−3 ± √31i )/4  

So, the roots x = (−3 + √31i )/4 and x = (−3 − √31i )/4.  where, i(iota) = √−1

Quadratic Equation Class 10^th in Hindi

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