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
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