- Quadratic Polynomial
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- Graphing quadratics Polynomial
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- what is a quadratic equation
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- 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
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- Nature of roots of Quadratic equation
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- Problem based on discriminant of a quadratic equation
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- Quadratic word problems

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

$ax^2 +bx+c=0$,

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 $b^2 -4ac > 0$, Quadratic equation has two real roots of different value

For $b^2 -4ac =0$, quadratic equation has one real root

For $b^2 -4ac < 0$, no real roots for quadratic equation

$3x^2 - 5x + 2 = 0$

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

and

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

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,

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

$16x - x^2 - 64 + 4x = 48$

$x^2 - 20x + 112 = 0$

Comparing this equation with $ax^2 + bx + c = 0$, we get

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

Discriminant = $b^2 - 4ac$

= (-20)

= 400 - 448 = -48

$b^2 - 4ac <0$ 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 $x^2 -6x=0$

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 $x^2 -16=0$

$x^2 -16=0$

$x^2 =16$

or$x=4 \; or \; -4$

Solve the quadratic equation by factorization method $x^2 -x -20=0$

$x^2-5x+4x-20=0$

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

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

or $x=-4 \; or \; 5$

Solve the quadratic equation by

For quadratic equation

$ax^2 +bx+c=0$,

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

$x=6 \; and \; -3$

- $x^{2} - 5 x - 14 = 0$
- $x^{2} - 3 x - 54 = 0$
- $x^{2} - 14 x + 45 = 0$
- $x^{2} - 8 x - 25 = -5$
- $x^{2} - 11 x + 18 = 0$
- $x^{2} - 13 x + 42 = 0$
- $>x^{2} + 3 x - 54 = 0$
- $x^{2} + 2 x +2 = 50$
- $x^{2} - 16 x + 63 = 0$
- $x^{2} - 15 x + 56 = 0$
- $x^{2} + 7 x + 18 = 48$

Class 10 Maths Class 10 Science

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