Notes on Quadratic equation | Grade 10 > Compulsory Mathematics > Algebra | KULLABS.COM

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An equation like ax2+ bx + c = 0 where a ≠ 0, which contains only one variable and '2' as its highest power is called a quadratic equation. There are two values of the variable in any quadratic equation. The roots of the equations are :

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

### Quadratic equation are of two types

e.g. x2 = 9 or, x2 - 9 = 0
i.e. ax2 + c = 0 ( a ≠ 0, c = 0 )

e.g. x2 - 9x - 15 = 0
i.e. ax2 + bx + c = 0 ( a ≠ 0, b ≠ 0)

An equation like ax2+ bx + c = 0 where a≠ 0, which contains only one variable and '2' as its highest power is called a quadratic equation. There are two values of the variable in any quadratic equation.. The roots of the equations are :

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

.

### Very Short Questions

Solution:

4x2 = 16x

or,4x2- 16x = 0

or, 4x ( x - 4) = 0

Either 4x = 0

∴ x = 0

Or, x - 4 = 0

∴ x = 4

Solution:

$$\frac{x^2+3}{2}$$ = 6

or, $$x^ 2 + 3$$ = 12

or, $$x^2$$ = 9

or, $$(x)^2$$ = (±3)2

or, x =±3

Let, the number be x. Hence, the other number is $$x^2$$

According to question,

x + $$x^2$$ = 20

or,x + $$x^2$$ - 20 = 0

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

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

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

Either, x - 4 = 0

∴ x = 4

Or, x + 5 = 0

∴ x = -5

Let, the three numbers be x, x + 1 and x + 2

Then, According to question,

$$x ^2$$ + $$(x+1)^2$$ + $$(x+2)^2$$ = 194

or, $$x ^2$$ + $$x ^2$$ + 2x + 1 + $$x ^2$$ + 4x + 4 = 194

or, 3$$x ^2$$ + 6x - 189 = 0

or, 3($$x ^2$$ + 2x - 63) = 0

or, ($$x ^2$$ + 2x - 63) = 0

or, $$x^2$$ + 9x - 7x - 63 = 0

or, x (x + 9) - 7( x+9) = 0

or, ( x - 7) ( x + 9) = 0

Either,

x - 7 = 0

∴ x = 7

Or, x + 9 = 0

∴ x = -9.

Therefore, the required numbers are 7, 7 + 1 , 7 + 2 = 7, 8, 9

Or, -9, -9 + 1, -9 + 2 = -9, -8, -7

Let, one of the odd number be x . Then, its consecutive odd number is ( x + 2)

According to question,

x ( x + 2) = 143

or, $$x^2$$ + 2x = 143

or, $$x ^ 2$$ + 2x - 143 = 0

or, $$x ^2$$ + 13x - 11x - 143 = 0

or, x ( x + 13) - 11( x + 13) = 0

or, ( x + 13) ( x - 11) = 0

Either,

x - 11 = 0

or, x = 11

Or, x + 13 =0

or, x = -13

Hence, the required number is 11 and 11 + 2 = 13

Let, one of the evennumber be x . Then, its consecutive evennumber is ( x + 2)

According to question,

x ( x + 2) = 168

or, $$x^2$$ + 2x = 168

or, $$x ^ 2$$ + 2x - 168 = 0

or, $$x ^2$$ + 14x - 12x - 168 = 0

or, x ( x + 14) - 12( x + 14) = 0

or, ( x + 14) ( x - 12) = 0

Either,

x - 12 = 0

or, x = 12

Or, x + 14 = 0

or, x = -14

Hence, the required number is 12 and 11 + 2 = 14

Let, one of the number be x . Then, its consecutive number is ( x + 1)

According to question,

x ( x + 1) =56

or, $$x^2$$ + x =56

or, $$x ^ 2$$ + x - 56= 0

or, $$x ^2$$ + 8x - 7x - 56= 0

or, x ( x + 8) - 7( x + 8) = 0

or, ( x + 8) ( x - 7) = 0

Either,

x - 7= 0

or, x =7

Or, x + 8=0

or, x = -8

Hence, the required number is 7 and 7 + 1=8

Let, the odd number be x and its consecutive odd number be ( x + 2).

According to question,

$$(x + 2) ^2$$ - $$x^2$$ = 24

or,$$x^2$$ + 4x + 4 -$$x^2$$ = 24

or, 4x = 24 - 4

or, 4x = 20

or, x = $$\frac {20}{4}$$

or, x = 5

Hence, the required numbers are 5 and 5 + 2 = 7

Let, the number be x.

Then, According to question,

4$$x ^ 2$$ = 16x

or 4 $$x ^2$$ - 16x = 0

or, 4x (x - 4) = 0

Either, 4x = 0

∴ x = 0

Or, x - 4 = 0

∴ x = 4.,

Let, the number be x.

Then, According to question,

$$x ^ 2$$ - 9 = 40

or,$$x ^ 2$$ = 40 + 9

or,$$x ^ 2$$ = 49

or, $$(x)^2$$ = $$(±7) ^ 2$$

∴x =±7

Let the whole number be x

Then, According to question,

x - 10 = 39× $$\frac {1}{x}$$

or, $$x ^ 2$$ - 10x = 39

or,$$x ^ 2$$ - 10x - 39 = 0

or,$$x ^ 2$$ - 13x + 3x - 39 = 0

or, x ( x - 13) + 3(x - 13) = 0

( x + 3) ( x - 13) = 0

Either x - 13 = 0

∴ x = 13

Or, x + 3 = 0

∴ x = - 3

Hence, the required whole number is 13.

Let, the natural number be x

Then,

According to quesiton,

$$\frac {x ^2}{2}$$ - 5 = 45

or $$\frac {x ^2}{2}$$ = 45 + 5

or,$$\frac {x ^2}{2}$$= 50

or, $$x ^ 2$$ = 100

or, $$( x)^2$$ = $$(10)^2$$ [ Only (+10) is taken because the required number is a natural number]

or, x = 10

Hence, the requierd number is 10.

Let, the first number is x then the second number becomes (12 - x)

According to question,

x ( 12 - x) = 32

or, -$$x ^ 2$$ + 12x - 32 = 0

or, $$x ^ 2$$ - 12x + 32= 0

or, $$x ^ 2$$ - 8x - 4x + 32 = 0

or, x ( x - 8) - 4( x - 8 ) = 0

or, ( x - 4) ( x - 8) = 0

Either, x - 4 = 0

∴ x = 4

Or, x - 8 = 0

∴ x = 8

If first number is 4, the other number is 12 - 4 = 8

If first number is 8, the second number is 12 - 8 = 4

Hence, the required numbers are 8,4 or 4,8

0%

6,6
6,4
6,5
6,7

34,35
43,36
43,34
43,33

9 and 9
9 and 7
9 and 8
none

6 and 5
6 and 3
6 and 4
6 and 2

6,7
all
6,6
6,8

13,17
12,17
12,18
none

11,3
11,4
11,5
all

15m, 9m
15m, 9m
14m , 9m
14m, 8m

9 cm, 12 cm
11 cm, 9cm
13cm, 9cm
9 cm , 11 cm

15
13
12
14
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##### Gyan

If two times the number exceeds,the number formed by interchanging its digit by four ???what does it mean ??