Question

Find the zeroes of the polynom.
4 x² +5root 2 x -3 and verify the relations
between zeroes and co-efficient of the
polynomials.


my confusion is in relationship of zeroes and coefficients. so plz write clearly .


any wrong and in-appropriate ans will be reported​

Answer 1

${\huge{\underline{\underline{\rm{Question}}}}}$

★ Find the roots of the Polynomial :

$\rm \: 4 {x}^{2} + 5 \sqrt{2} \: x - 3 = 0$

And verify the relationship between zeroes and coefficient of the Polynomial .

${\huge{\underline{\underline{\rm{Answer\checkmark}}}}}$

First , we will find the zeroes of the Polynomial .

$\rm \: 4 {x}^{2} + 5 \sqrt{2} \: x - 3 = 0$

Standard equation (2°) is in the form of -

$\rm \: a {x}^{2} + bx + c = 0$

So ,

In the Polynomial ,

• a = 4
• b = $5\sqrt{2}$
• c = -3

Now using the Formula method , finding the zeroes

$\rm \mapsto \: roots \: = \dfrac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

$\mapsto \rm \: roots = \dfrac{ - 5 \sqrt{2} \pm \sqrt{ {(5 \sqrt{2}) }^{2} - 4 \times 4 \times ( - 3) } }{2 \times 4}$

$\rm \mapsto \: roots \: = \dfrac{ - 5 \sqrt{2} \pm \sqrt{50 + 48} }{8}$

$\rm \mapsto \: roots \: = \dfrac{ - 5 \sqrt{2} \pm \sqrt{98} }{8}$

$\rm \mapsto \:roots = \dfrac{ - 5 \sqrt{2} \pm \sqrt{49 \times 2} }{8}$

$\rm \mapsto \: roots = \dfrac{ - 5 \sqrt{2} \pm7 \sqrt{2} }{8}$

$\mapsto \rm \: first \: zero \: = \dfrac{ - 5 \sqrt{2} + 7 \sqrt{2} }{8}$

$\mapsto \boxed{\rm first \: zero \: = \frac{ \cancel2 \sqrt{2} }{ \cancel8} = \dfrac{ \sqrt{2} }{4} = \frac{1}{2 \sqrt{2} } }....(i)$

$\rm \mapsto \: second \: zero = \dfrac{ - 5 \sqrt{2} - 7 \sqrt{2} }{8}$

$\mapsto \boxed{ \rm \: second \: zero = \dfrac{ - \cancel{12} \sqrt{2} }{ \cancel8} = \frac{ - 3 \sqrt{2} }{2} = \frac{ - 3}{ \sqrt{2} } }$

_________________________

Now , relationship ..

as we know ,

• Sum of zeroes = $\rm\dfrac{-b}{a}$

Here ,

$\rm \:first \: zero + second \: zero \: = \dfrac{ - 5 \sqrt{2} }{4}$

LHS ,

$\rm \: 1st \: zero + 2nd \: zero \: = \frac{1}{2 \sqrt{2} } + \frac{ - 3}{ \sqrt{2} } \\ \\ \implies \: \frac{1 - 3 \times 2}{2 \sqrt{2} } = \frac{1 - 6}{2 \sqrt{2} } = \boxed{\frac{ - 5}{2 \sqrt{2} } }$

RHS ,

$\dfrac{ - 5 \sqrt{2} }{4} = \dfrac{ - 5 \cancel{\sqrt{2} }}{2 \sqrt{2} \times \cancel{\sqrt{2} } } = \boxed{ \dfrac{ - 5}{2 \sqrt{2} } }$

Since ,

LHS = RHS ,

The relation between zeroes of the Polynomial and coefficient is established .

Answer 2

Step-by-step explanation:

the above answer is absolutely correct