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Sia ? 6 years, 4 months ago
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Posted by Sakshi Agarwal 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
We have 2x2 - 7x + 3 = 0
{tex}\implies2( x^2 - {7 \over 2}x + {3\over 2}) = 0{/tex}
{tex}\implies x^2 - {7 \over 2}x + {49 \over 16} = {-3 \over 2} +{ 49 \over 16}{/tex} (Adding 49/16 to both sides)
{tex}\implies x^2 -2 \times x \times {7 \over 4} + ({7 \over 4})^2 = {-24 +49 \over 16}{/tex}
{tex}\implies (x-{7\over4})^2 = {25 \over 16}{/tex}
{tex}\implies x-{7\over 4}= \pm \sqrt({25 \over 16}){/tex}
{tex}\implies x={7\over 4} \pm {5 \over 4}{/tex}
{tex}\implies x={7\over 4} + {5 \over 4}\, and \,x={7\over 4} - {5 \over 4}{/tex}
{tex}\implies x=3\, and \,{1\over 2}{/tex}
{tex}\therefore{/tex}the roots of the given equation are {tex}3{/tex} and {tex}1\over 2{/tex}.
Posted by Harsh Dhariwal 7 years, 4 months ago
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Sanjivani Gawade 7 years, 4 months ago
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Shashank Mishra 7 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let a be any positive integer and b = 6
∴ by Euclid’s division lemma
a = bq + r, 0≤ r and q be any integer q ≥ 0
∴ a = 6q + r,
where, r = 0, 1, 2, 3, 4, 5
If a is even then then remainder by division of 6 is 0,2 or 4
Hence r=0,2,or 4
or A is of form 6q,6q+2,6q+4
As, a = 6q = 2(3q), or
a = 6q + 2 = 2(3q + 1), or
a = 6q + 4 = 2(3q + 2).
If these 3 cases a is an even integer.
but if the remainder is 1,3 or 5 then r=1,3 or 5
or A is of form 6q+1,,6q+3 or,6q+5
Case 1:a = 6q + 1 = 2(3q) + 1 = 2n + 1,
Case 2: a = 6q + 3 = 6q + 2 + 1,
= 2(3q + 1) + 1 = 2n + 1,
Case 3: a = 6q + 5 = 6q + 4 + 1
= 2(3q + 2) + 1 = 2n + 1
This shows that odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Posted by Student Commerce 7 years, 4 months ago
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Posted by Himanshu Sharma 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let the ages of A and B be x and y years respectively.
Then, age difference of A and B is 2 years.
{tex}\Rightarrow{/tex}x - y = {tex}\pm{/tex}2
A's father D is twice as old as A and B is twice as old as his sister C.
Then, D's age =2x years. and, C's age {tex}= \frac { y } { 2 }{/tex}years.
Clearly, D is older than C
The age of D and C differ by 40 years.
{tex}\therefore{/tex} {tex}2 x - \frac { y } { 2 } = 40 {/tex}
{tex}\Rightarrow 4 x - y = 80{/tex}
Thus, we have the following two systems of linear equations
x - y =2 ...............(i)
and, 4x - y =80 ...........(ii)
x - y = -2 ............(iii)
and, 4x - y=80 ..................(iv)
Subtracting equation (i) from equation (ii), we get
(x - y) - (4x - y)= 2 - 80
⇒ x - y - 4x + y = -78
⇒ - 3x = - 78
⇒ 3x =78
⇒x =26
Putting x =26 in equation (i), we get
x - y =2
⇒ 26 - y = 2
⇒ y =24
Subtracting equation (iv) from equation (iii), we get
{tex}- 3 x = - 82 {/tex}
{tex}\Rightarrow x = \frac { 82 } { 3 } = 27 \frac { 1 } { 3 }{/tex}
Putting {tex}x = \frac { 82 } { 3 }{/tex}in equation (iii), we get
{tex}y = \frac { 82 } { 3 } + 2 = \frac { 88 } { 3 } = 29 \frac { 1 } { 3 }{/tex}
Hence, A's age =26 years and B's age =24 years
or,
A's age {tex}= 27 \frac { 1 } { 3 }{/tex}Years and B's age {tex}29 \frac { 1 } { 3 }{/tex}years.
Posted by Rahul Singh 7 years, 4 months ago
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Sia ? 6 years, 5 months ago
Putting {tex}\frac 1x{/tex}= u and {tex}\frac 1y{/tex} = v, the given equations become
{tex}2u+ 3v = 13{/tex} .... (i)
{tex}5u - 4v = -2{/tex} .......(ii)
{tex}\text{Multiplying (i) by 4 and (ii) by 3 and adding the results, we get}{/tex}
{tex}8u + 15u = 52 - 6{/tex}
{tex}\Rightarrow{/tex} {tex}23u = 46{/tex}
{tex}\Rightarrow \quad u = \frac { 46 } { 23 } = 2{/tex}
{tex}\text{Putting u = 2 in (i), we get}{/tex}
(2 {tex}\times {/tex} 2) + 3v = 13 {tex}\Rightarrow{/tex} 3v = 13 - 4 = 9 {tex}\Rightarrow{/tex} v = 3.
Now,u = 2{tex}\Rightarrow \frac { 1 } { x } = 2 \Rightarrow 2 x = 1 \Rightarrow x = \frac { 1 } { 2 }{/tex}
And, v = 3 {tex}\Rightarrow \frac { 1 } { y } = 3 \Rightarrow 3 y = 1 \Rightarrow y = \frac { 1 } { 3 }{/tex}
Hence, {tex}x = \frac { 1 } { 2 } \text { and } y = \frac { 1 } { 3 }{/tex}
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Sagar Menaria 7 years, 4 months ago
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