- Quadratic Polynomial
- |
- Graphing quadratics Polynomial
- |
- what is a quadratic equation
- |
- How to Solve Quadratic equations
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- Factoring quadratics equations
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- Solving quadratic equations by completing the square
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- Solving quadratic equations by using Quadratic formula
- |
- Nature of roots of Quadratic equation
- |
- Problem based on discriminant of a quadratic equation
- |
- Quadratic word problems

- NCERT Solutions Quadratic Equation Exercise 1
- |
- NCERT Solutions Quadratic Equation Exercise 2
- |
- NCERT Solutions Quadratic Equation Exercise 3
- |
- NCERT Solutions Quadratic Equation Exercise 4

For quadratic equation

ax

roots are given by

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

and

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

This has been derived using the square method as discussed above only

For b

For b

For b

3x

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

and

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

b

S.no |
Condition |
Nature of roots |

1 |
b ^{2} -4ac > 0 |
Two distinct real roots |

2 |
b ^{2}-4ac =0 |
One real root |

3 |
b ^{2}-4ac < 0 |
No real roots |

The sum of the ages of two friends A and B is 20 years. Four years ago, the product of their ages in years was 48.

Here we need to advise if that is a possible condition

Let the age of A be x years.

then the age of the B will be (20 - x) years.

Now 4 years ago,

Age of A = (x - 4) years

Age of N = (20 - x - 4) = (16 - x) years

So we get that,

(x - 4) (16 - x) = 48

16x - x

x

Comparing this equation with ax2 + bx + c = 0, we get

a = 1, b = -20 and c = 112

Discriminant = b

= (-20)

= 400 - 448 = -48

b2 - 4ac < 0 Therefore, there will be no real solution possible for the equations. Such type of condition doesn't exist

Is the following statement 'True' or 'False'?Justify your answer.

If in a quadratic equation the coefficient of x is zero, then the quadratic equation has no

real roots

False, since the discriminant in this case is -4ac which can still be nonnegative if a and c are of opposite signs or if one of a or c is zero.

Find the roots of the quadratic equation x

There is no constant term in this quadratic equation, we can x as common factor

x(x-6)=0

So roots are x=0 and x=6

Solve the quadratic equation x

x

x

or x=4 or -4

Solve the quadratic equation by factorization method x

x

x(x-5)+4(x-5)=0

(x+4)(x-5)=0

or x=-4 or 5

Solve the quadratic equation by

For quadratic equation

ax

roots are given by

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

and

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

Here a=1

b=-3

c=-18

Substituting these values,we get

x=6 and -3

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