- 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 4.2
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- NCERT Solutions Quadratic Equation Exercise 4.2
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- NCERT Solutions Quadratic Equation Exercise 4.3
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- NCERT Solutions Quadratic Equation Exercise 4.4

Given below are the

- Concepts questions
- Calculation problems
- Multiple choice questions
- Long answer questions
- Fill in the blank's

State which all quadratic equations have real roots, no real roots

- $x^2 + x+7=0$
- $3x^2 +6x+1=0$
- $9x^2 +x +3=0$
- $11x^2 -12x-1=0$
- $-13x^2 +3x+7=0$
- $2x^2 -6x+3=0$
- $x- \frac {1}{x}-3=0$, x≠0
- $-x^2 -2x-2=0$

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 |

No real roots : (a) ,(c),(h)

Find the roots of the quadratic equation using factorization technique

a. $x^2-3x-10=0$

b. $x^2 -11x+30=0$

a.

$x^2-3x-10=0$

$x(x-5) +2(x-5)=0$

$(x+2)(x-5)=0$

So roots are x=-2 and 5

b. Roots are 5 and 6

Find the roots of the quadratic equation using square method

a. $x^2 +4x-5=0$

b. $2x^2-7x+3=0$

a.

$(x+ \frac {4}{2})^2 -(\frac {4}{2})^2 -5=0$

$(x+2)^2-9=0$

$(x+2)^2=9$

$x+2=\pm 3$

x=1 or -5

b.

$(x-\frac {7}{4})^2 -(\frac {7}{4})^2 +3/2=0$

$(x-\frac {7}{4})^2= \frac {49}{16} - \frac {3}{2}=0$

$(x-\frac {7}{4})^2=\frac {25}{16}$

$x-\frac {7}{4}=\pm \frac {5}{4}$

or

x=1/2 or 3

a. There are no reals roots of the quadratic equation $x^2+4x+5=0$

b. The roots of the equation $x^2-1=0$ are 1,-1

c. A quadratic equation can have at most 2 real roots

d. In a quadratic equation $ax^2 +bx+c=0$ ,if a and c are of same sign and b is zero ,the quadratic equation has real roots

e. In a quadratic equation $ax^2 +bx+c=0$ ,if a and c are of opposite sign, then quadratic equation will definitely have real roots

f. for k > 0,the quadratic equation $2x^2+6x-k=0$ will definitely have real roots

g. if the roots of the quadratic equation are rational, the coefficient of the term x will also be rational.

h. if the roots of the quadratic equation are irrational, the coefficient of the term x will also be irrational

i. Every quadratic equation will have rational roots

- True
- True
- True
- false
- True
- True
- true
- true
- False

Find a natural number whose square diminished by 84 is thrice the 8 more of given number

a. 21

b. 13

c.11

d. 12

$x^2 -84=3(x+8)$

$x^2-3x-108=0$

x= 12 or -9

So answer is 12

The roots of the quadratic equation

$x^2 +14x+40=0$ are

a.(4,10)

b.(-4,10)

c.(-4,-10)

d.(4,-10)

$x^2 +14x+40=0$

$x^2 + 4x + 10x +40=$

$x(x+4) + 10(x+4)=0$

$(x+4)(x+10)=0$

x=-4 or -10

Answer (c)

The equation

$x^5 +x+20=0$

a. is a quadratic equation

b. is not a quadratic equation

It is not a quadratic equation

The roots of the quadratic equation

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

a. are real

b. are not real

$b^2 - 4ac = 4 -20 =-16$

Answer is (b)

Which one of the following is not a quadratic equation?

a. $(x + 2)^2= 2(x + 3)$

b. $x^2 + 3x = (-1) (1 - 3x)^2$

c. $(x + 2) (x - 1) = x^2 - 2x - 5$

d. $x^3 - x^2 + 2x + 1 = (x + 1)^3$

We can expand each of these expression and compare with $ax^2 + bx+x=0$.

Answer (c)

Find the roots of

$6x^2- \sqrt {2}x - 2 = 0$ by the factorisation of the corresponding quadratic polynomial.

a. $-\frac {\sqrt {2}}{3},\frac {\sqrt {2}}{2}$

b. $-\frac {\sqrt {1}}{3},\frac {\sqrt {2}}{2}$

c. $-\frac {\sqrt {2}}{3},\frac {\sqrt {2}}{5}$

d. None of the these

$6x^2- \sqrt {2}x - 2 = 0$

$6x^2 + 3 \sqrt {2}x -2\sqrt {2}x - 2 = 0$

$ 3x(2x - \sqrt {2}) - \sqrt {2}(2x - \sqrt {2})=0$

$(3x- \sqrt {2})(2x - \sqrt {2})=0$

or x = $-\frac {\sqrt {2}}{3},\frac {\sqrt {2}}{2}$

Answer (a)

Had Ram scored 10 more marks in her mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test?

a. 15 marks

b. 10 marks

c. 12 marks

d. 20 marks

$9 (x +10) = x^2$

$x^2 - 9x - 90 = 0$

$(x + 6) (x -15) = 0$

Therefore, x = - 6 or x =15

Answer (a)

Match the roots with the quadratic equations

b-> iii, vii

c -> iii, v

d -> iv, viii

e. i, ii

Given below are the links of some of the reference books for class 10 math.

- Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
- Mathematics for Class 10 by R D Sharma
- Pearson IIT Foundation Maths Class 10
- Secondary School Mathematics for Class 10
- Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.

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