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Ask QuestionPosted by Yashif Alam 7 years, 6 months ago
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Posted by Shaaa Kukkarni 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
The given quadratic equation is
{tex}3 x ^ { 2 } - 4 \sqrt { 3 } x + 4 = 0{/tex}
Here, a = 3, {tex}b = - 4 \sqrt { 3 }{/tex}, c = 4
{tex}\therefore{/tex} discriminant = b2 - 4ac
{tex}= ( - 4 \sqrt { 3 } ) ^ { 2 } - 4 ( 3 ) ( 4 ){/tex}
= 48 - 48 = 0
Hence, the given quadratic equation
has two equal real roots.
The roots are {tex}= - \frac { b } { 2 a } , - \frac { b } { 2 a }{/tex}
{tex}\text { i.e. } - \frac { ( - 4 \sqrt { 3 } ) } { 2 \times 3 } , - \frac { ( - 4 \sqrt { 3 } ) } { 2 \times 3 } , \text { i.e. } \frac { 2 } { \sqrt { 3 } } , \frac { 2 } { \sqrt { 3 } }{/tex}
Posted by Harshita Agnihotri 7 years, 6 months ago
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Rupesh Kumar 7 years, 6 months ago
Posted by Sunny Yadav 7 years, 6 months ago
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Posted by Sujal Choudhary 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
n3 - n = n (n2 - 1) = n (n - 1) (n + 1)
Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.
If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 3.
⇒ n (n – 1) (n + 1) is divisible by 3.
Similarly, whenever a number is divided by 2, the remainder obtained is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q, then n is divisible by 2.
If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.
⇒ n (n – 1) (n + 1) is divisible by 2.
Since, n (n – 1) (n + 1) is divisible by 2 and 3.
∴ n (n-1) (n+1) = n3 - n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)
Posted by Ammu Ammu 7 years, 6 months ago
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Samit Gautam 7 years, 6 months ago
Susmita Mandal 7 years, 6 months ago
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Aditi Verma 7 years, 6 months ago
Posted by Abhishek Yadav 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
{tex}\triangle ADE \sim \triangle ABC{/tex}
{tex}\Rightarrow \frac{ar(\triangle ADE)}{ar(\triangle ABC)}=(\frac{DE}{BC})^2{/tex}
{tex}\Rightarrow \frac{{45}}{ar(\triangle ABC)}=(\frac{6}{8})^2{/tex}
{tex}\Rightarrow \frac{{45}}{ar(\triangle ABC)}=\frac{36}{64}{/tex}
{tex}\Rightarrow ar(\triangle ABC)=\frac{64(45)}{36}{/tex}
{tex}\Rightarrow ar(\triangle ABC)=80 cm^2{/tex}
Posted by Dipanshi Gupta 7 years, 6 months ago
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M B 7 years, 6 months ago
Posted by Ujjwal Malhotra 7 years, 6 months ago
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Posted by Gourav Chaurasiya 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer.
By Euclid’s division lemma, we have
a = bq + r; 0 ≤ r < b
For a = n and b = 3, we have
n = 3q + r ...(i)
Where q is an integer and 0 ≤ r < 3, i.e. r = 0, 1, 2.
Putting r = 0 in (i), we get
{tex}n = 3q{/tex}
∴ n is divisible by 3.
{tex}n + 1 = 3q + 1{/tex}
∴ n + 1 is not divisible by 3.
{tex}n + 2 = 3q + 2{/tex}
∴ n + 2 is not divisible by 3.
Putting r = 1 in (i), we get
{tex}n = 3q + 1{/tex}
∴ n is not divisible by 3.
{tex}n + 1 = 3q + 2{/tex}
∴ n + 1 is not divisible by 3.
{tex}n + 2 = 3q + 3 = 3(q + 1){/tex}
∴ n + 2 is divisible by 3.
Putting r = 2 in (i), we get
{tex}n = 3q + 2{/tex}
∴ n is not divisible by 3.
{tex}n + 1 = 3q + 3 = 3(q + 1){/tex}
∴ n + 1 is divisible by 3.
{tex}n + 2 = 3q + 4{/tex}
∴ n + 2 is not divisible by 3.
Thus for each value of r such that 0 ≤ r < 3 only one out of n, n + 1 and n + 2 is divisible by 3.
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Sia ? 6 years, 6 months ago
Check NCERT solutions here : <a href="https://mycbseguide.com/ncert-solutions.html">https://mycbseguide.com/ncert-solutions.html</a>
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Shweta Yadav 7 years, 6 months ago
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Aarohi Raichand 7 years, 6 months ago
7Thank You