English, asked by gopikakanakan, 11 hours ago

Activity 3( Comparison)

you are aware of the pivotal role of a woman , especially the mother at home .there are occasions when her self- effacing love is painfully missing ,when the prayer Take me not till the children grow!has not been realised.​

Answers

Answered by golusingh444456
0

Explanation:

\large\underline{\sf{Solution-}}

Solution−

Let assume that

\rm \: \alpha , \beta \: are \: zeroes \: of \: polynomial \: {2x}^{2} + px + 45α,βarezeroesofpolynomial2x

2

+px+45

We know,

\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}

Sum of the zeroes=

coefficient of x

2

−coefficient of x

\rm\implies \: \alpha + \beta = - \: \dfrac{p}{2}⟹α+β=−

2

p

Also, we know that

\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}

Product of the zeroes=

coefficient of x

2

Constant

\rm\implies \: \alpha \beta = \dfrac{45}{2}⟹αβ=

2

45

Now, According to statement,

\rm \: {( \alpha - \beta )}^{2} = 144(α−β)

2

=144

can be rewritten as

\rm \: {( \alpha + \beta )}^{2} - 4 \alpha \beta = 144(α+β)

2

−4αβ=144

\bf \:\bigg [ \: \because \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy \: \bigg][∵(x+y)

2

−(x−y)

2

=4xy]

On substituting the values from above, we get

\rm \: {\bigg[ - \dfrac{p}{2} \bigg]}^{2} - 4 \times \dfrac{45}{2} = 144[−

2

p

]

2

−4×

2

45

=144

\rm \: \dfrac{ {p}^{2} }{4} - 90 = 144

4

p

2

−90=144

\rm \: \dfrac{ {p}^{2} }{4} = 144 + 90

4

p

2

=144+90

\rm \: \dfrac{ {p}^{2} }{4} = 234

4

p

2

=234

\rm \: {p}^{2} = 4 \times 234p

2

=4×234

\rm \: p= \pm \: \sqrt{4 \times 234}p=±

4×234

\rm \: p= \pm \: \sqrt{2 \times 2 \times2 \times 3 \times 3 \times13 }p=±

2×2×2×3×3×13

\rm\implies \:\rm \: p= \pm \: 6\sqrt{26}⟹p=±6

26

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ADDITIONAL INFORMATION :-

For Cubic Polynomial

\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}:⟼α,β,γarezeroesofax

3

+bx

2

+cx+d,then

\begin{gathered}\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}} \\ \end{gathered}

α+β+γ=−

a

b

\begin{gathered}\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}} \\ \end{gathered}

αβ+βγ+γα=

a

c

\begin{gathered}\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}} \\ \end{gathered}

αβγ=−

a

d

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