# Quadratic equation for class 10 board 2020-21

The zeroes of polynomial f(x) are called roots of equation f(x)=0. We know the degree of the quadratic equation is 2 .So, it can have at most two zeroes which may be real or imaginary, it depends on discriminant which we  will discuss later in this post

Let’s learn to identify quadratic equation through examples

Here are a few examples

(i) x²-6x+4=0

(ii)2x²-7x=0

(iii)x+3/x=x²

(iv)x²+2√x+5=0

solution(i) The degree of the equation is 2 and the power of every variable x of the equation is non-fractional. so, it is a quadratic equation.

solution(ii) The degree of the equation is 2 and the power of every variable x of the equation is non-fractional. so, it is a quadratic equation.

solution(iii) Here, the highest power of the variable is 2 but the power of a variable x is negative(-1).so it is not a quadratic equation

solution(iv) Again highest power of variables is 2 but the power of a variable is x is fractional (½).so it is also not a quadratic equation.

### Determine whether the given value is the solution of the quadratic equation or not.

It is so simple to find whether a value is the solution of the quadratic equation or not.  Let α is any real number and f(x)=0 is a quadratic equation if f(α)=0 then α is a solution of equation otherwise not .understand it with an example

Example: 3x²-2x-1=0 is a quadratic equation, check whether 1 is a solution of an equation or not?

solution: substitute x=1 on the LHS of the equation ,we get

3(1)²-2(1)-1=3-2-1

=3-3

=0

so,x=1 is a solution of the equation

### On determining an unknown variable in a quadratic equation when its roots are given

Example: Determine the value of K for which the given value is a solution of the equation

(i)kx²+2x-3=0,x=2

(ii) x²+2ax-k=0,x=-a

solution(i) Since we have 2 as a root of the equation

∴                       f(2)=0

k(2)²+2(2)-3=0

4k +4 -3=0

4k+1=0

k=-1/4

solution(ii) Since we have -a as a root of the equation

∴                              f(-a)=0

(-a)²+2a(-a)-k=0

a²-2a²-k=0

-k=a²

k=-a²

### Quadratic equation class 10 notes :[formulation of the quadratic equation ]

Q.1 The product of two consecutive positive integers is 240 .formulate the quadratic equation whose roots are these integers.

solution(i) Let two consecutive positive integers be x and x+1.

since the product of these integers is 240

x(x+1)=240

x²+x=240

x²+x-240=0

This is our required quadratic equation

Q.2 Rohan’s mother is 26 years older than him. The product of their ages 3 years from now will be 360.formulate the quadratic equation to find their ages.

solution(ii) Let the present age of Rohan be x years .then, their mother’s age will be (x+26)years.

After 3 years Rohan’s and his mother age will be (x+3) and (x+26+3) years respectively.

It is given that the product of their ages after 3 years from now is 360

∴                        (x+3)(x+29)=360

(x²+29x+3x+87)=360

(x²+32x+87)=360

x²+32x =360-87

x²+32x=273

x²+32x-273=0

This is required equation.

we will later practice more question on this

## The solution of a quadratic equation by factorization.

Q. Solve the following quadratic equation by factorization.

(i)x²+6x+5=0

(ii)1/x-2+2/x-1=6/x

(iii)x²+2√2x-6=0

solution(i)we have x²+6x+5=0

Factor of constant term=5=1×5

Express factor of the constant term in such a way that sum or difference is the middle term and product is the constant term

x²+5x+x+5=0

x(x+5)+(x+5)=0

(x+5)(x+1)=0

for zeroes     (x+5)=0 or (x+1)=0

x=-5 and x=-1

solution(ii) we have ,

1/x-2+2/x-1=6/x

when we solve it taking LCM it will convert into 3x²-13x+12=0

3x²-13x+12=0

Factor of constant term =3×12            (we consider the coefficient of x² as a constant term)                                                   =3×3×2×2×1

Express factor of the constant term in such a way that their sum or difference is the middle term and product is the constant term

3x²-9x-4x+12=0

3x(x-3)-4(x-3)=0

(x-3)(3x-4)=0

For zeroes

(x-3)=0   or      (3x-4)=0

x=3                       x=4/3

solution(iii) x²+2√2x-6=0

Factor of constant term=6=1×2×3.but middle term is in the form of square root √2 so,we have to convert 2 of factor into √2.√2.then ,continue same process

x²+2√2x-6=0

x²+3√2x-√2x-6=0

x(x+3√2)-√2(x+3√2)=0

(x+3√2)(x-√2)=0

for zeroes

(x+3√2)=0   or    (x-√2)=0

x=-3√2    or        x=√2

## Steps to proof quadratic formula(completing the square method)

Proof: we know the general form of quadratic equation is ax²+bx+c =0 where a≠0

step:1 Make the coefficient of x²unity(1) by dividing throughout by a.

\frac{ax^{2} +bx+c}{a} =\frac{0}{a}=x^{2} +\frac{bx}{a} +\frac{c}{a} =0

step:2 Shift the constant term on RHS

x^{2} +\frac{bx}{a} =-\frac{c}{a}

step:3  Add square of half of the coefficient of x i.e (b/2a)²on both sides to obtain

x^{2} +2\frac{bx}{2a} +\left(\frac{b}{2a}\right)^{2} =\left(\frac{b}{2a}\right)^{2} +\left( -\frac{c}{a}\right)

step:4 write LHS as the perfect square of a binomial expression and simply RHS to get

\left( x+\frac{b}{2a}\right)^{2} =\frac{b^{2} -4ac}{4a^{2}}

then it will become

x= \frac{-b\pm \sqrt{b^{2} -4ac}}{2a}

Q.1 Solve the equation 2x²-5x+3=0 by the method of completing the square.

solution(i) we have ,  2x²-5x+3=0

2x²-5x+3/2=0    (divinding both side by 2)

x²-5x/2=-3/2     (shifting the constant term on RHS)

Add the square of half of the coefficient of x on both side

so, we have -5/2 as the coefficient of x .It’s half square become (-5/4)²

x²-2.5x/4+(-5/4)²=-3/2+(-5/4)²

(x-5/4)²=-3/2+25/16

(x-5/4)²=1/16

x-5/4= ±1/4

we shall get x=3/2 and x=1 by solving taking positive and negative sign.

Q.2 By using the method of completing the square ,show that the equation 4x²+3x+5=0 has no real roots.

sol.we have ,

4x²+3x+5=0

Divide both side by 4

\frac{4x^{2} +3x+5}{4} =\frac{0}{4}

it become

x²+\frac{3x}{4}+\frac{5}{4}=0

shift the constant term on RHS

x²+(\frac{3x}{4})=-\frac{5}{4}

Now add half of square of coefficient of x on both side

x²+(\frac{3x}{4})+(\frac{3}{8})²=(\frac{3}{8})²-\frac{5}{4}

After simplifying it become

(x+\frac{3}{8})²=-\frac{71}{64}

here RHS is negative .But LHS cannot be negative because square of any number is always positive

so given equation has no real roots

Q.3 Solve the equation x²-(√3+1)x+√3=0 by the method of completing the square .

sol. we have x²-(√3+1)x+√3=0 but coefficient of x² is 1 so we needn’t to divide

simply transfer constant term on RHS

x²-(√3+1)x=-√3

Add square of half of coefficient of x on both side

x²-(√3+1)x+(-\frac{√3+1}{2})²=-√3+(-\frac{√3+1}{2})²

After simplifying

(x-\frac{√3+1}{2})²=\frac{-4√3+( √3+1)²}{4}  (x-\frac{√3+1}{2})²= (\frac{√3-1}{2})² x-\frac{√3+1}{2}=  ±\frac{√3-1}{2}  x=  \frac{√3+1}{2}±\frac{√3-1}{2}

solve taking positive and negative sign respectively we get,

x=√3,1

## The solution by quadratic formula (shreedharacharya’s formula)

we cannot solve all quadratic equation by factorisation such as those quadratic equation whose zeroes are nonreal .In that cases we solve quadratic equation with the help of quadratic formula (shreedharacharya’s formula)

which prove we learned earlier in completing the square.

There is one very important term which you must know i.e Discriminant.This helps us to know whether the roots of the equation are real or nonreal

### Quadratic equation class 10 notes :[Discriminant]

Discriminant: If ax²+bx+c=0, a≠0is a quadratic equation, then the expression b²-4ac is known as its discriminant and generally denoted by D

so we can write Quadratic formula in the form of discriminant as

x=\frac{-b\pm \sqrt{D}}{2a}

Q.1 Determine whether the given equation has real roots and if so, find the roots.
(i)9x²+7x-2=0

sol(i) The given equation is 9x²+7x-2=0and we can only determine whether its roots are real or nonreal by Discriminant i.e D=b²-4ac

here  a=9 ,b=7 and c=-2

∴      D=(7)²-4(9)(-2)

D=49+72

D=121

here D value is greater than o then equation has real roots

Now we have to find its roots

x=\frac{-b\pm \sqrt{D}}{2a}

substitute a=9,b=7 and D=121 in above equation

x =\frac{-7\pm \sqrt{121}}{29}

solve taking positive and negative sign respectively

The roots of given equation are \frac{2}{9} and -1

### Nature of roots

we can determine the roots of any equation whether it is real or nonreal by finding its Discriminant.

Case.1 When the value of D i.e b²-4ac is greater than 0 D>0

roots of the equation are real

case.2 when the value of D<0

roots of the equation are nonreal

case.3 when the value of D=o

then, both roots of the equation are real and equal

let’s take about some magical things which you can try

This is a type of calculator which solves quadratic formula (find its roots). What you have to do with that

Simply compare your  equation which you want to solve with ax²+bx+c =0 where a≠0

and find the value of a,b, and c in the calculator.The calculator will automatically give you the roots of the equation.

#### Completing the square calculator

In this calculator, if you enter the value of a,b and c .It will automatically give you the solution but the solution will explain in the format of completing the square. You should also try this

#### Factoring calculator

I think you have a guess about this calculator, if not I explain.Do you want to know the factor of any equation,you can immediately find that as well and if you also excited to know the solution, you can find that as well.Here is the link to the factoring calculator

If you enjoyed these calculators,don’t forget to comment