Question

if m and n are zeroes of the polynomial 3x2+11x+4 then find the value of a) m+n b) m×n c) 1/m+1/n d) m2+n2
Plz it is urgent .....
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Answer 1

Quadratic Equation:-

x²+11x+4

Given:-

Roots = m and n

To Find:-

1) m+n

2) mxn

3) 1/m +1/n

4) m²+n2

Solution:-

Concept,

Sum of roots = α+β= -b/a

Product of Roots= αβ= c/a

Using these two concepts,

1) m+n = -b/a

=> -11/3

2)mxn = c/a

=> 4/3

3) 1/m+1/n

=> (m+n)/mn

=> m+n= -11/3

and mn = 4/3

=> So,

-11/3 / 4/3

=> 1/m+1/n = -11/4

4) m²+n²

=> (m+n)²-2mn

=> (-11/3)² -2(4/3)

=> 121/9 -8/3

=> => 121-24/9

=> 97/9

Ans

Answer 2

➷➷➷➷➷ＱＵＥＳＴＩＯＮ➷➷➷➷➷

If m and n are zeroes of the polynomial

$\mathtt{3 {x}^{2} + 11x + 4}$

then find the value of

$\mathtt{a)m + n} \\ \\ \mathtt{b)m \times n} \\ \\ \mathtt{c) \frac{1}{m} + \frac{1}{n}} \\ \\ \mathtt{ {m}^{2} + {n}^{2} }$

➷➷➷➷ＳＯＬＵＴＩＯＮ➷➷➷➷

Comparing the given quadratic polynomial with the standard form of quadratic polynomial

$\tt{3 {x}^{2} + 11x + 4}$

and

$\tt{a {x}^{2} + bx + c}$

we will get ,

a = 3 ; b = 11 ; c = 4

Now as we know,

sum of zeroes = - b / a

so,

m + n = - 11 / 3

$\tt{\boxed{m + n = \frac{ - 11}{3} }}$

,

and

product of zeroes = c / a

so,

m × n = 4 / 3

$\tt{\boxed{m \times n = \frac{4}{3} }}$

Now we have to find

$\tt{ \frac{1}{m} + \frac{1}{n} } \\ \\ \tt{ = \frac{m + n}{mn}} \\ \\ \tt{(putting \: values \: from \: above \: answers)} \\ \\ \tt{ = \frac{ - 11}{3} \div \frac{4}{3} = \frac{ - 11}{3} \times \frac{3}{4} } \\ \\ \tt{ = \frac{ - 11}{4} } \\ \\ \tt{hence} \\ \\ \tt{\boxed{ \frac{1}{m} + \frac{1}{n} = \frac{ - 11}{4} }}$

next, have to find

$\tt {m}^{2} + {n}^{2}$

,

$\tt {m}^{2} + {n}^{2} = {(m + n)}^{2} - 2mn \\ \\ \tt \:( putting \: values) \\ \\ = {( \frac{ - 11}{3}) }^{2} - 2( \frac{4}{ 3} ) \\ \\ \tt = \frac{121}{9} - \frac{8}{3} = \frac{121 - 24}{9} \\ \\ \tt = \frac{97}{9} \\ \\ \tt \: hence \\ \\ \tt{\boxed{ {m}^{2} + {n}^{2} = \frac{97}{9} }}$