I want question answer
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class 6
Answers
Answer:
Given is an Equation:
\sf{}( {c}^{2} - ab) {x}^{2} - 2( {a}^{2} - bc)x + {b}^{2} - ac = 0(c
2
−ab)x
2
−2(a
2
−bc)x+b
2
−ac=0
By comparing given equation with general form of a Quadratic equation which is ;
\implies \sf{}a {x}^{2} + bx + c = 0⟹ax
2
+bx+c=0
Here , a = (c² - ab)
b = -2 ( a² - bc )
c = b² - ac
Now , As it is given that roots of given equation are real and equal .Therefore ,
\implies \sf{}D = 0⟹D=0
By using Discriminant formula ;
\implies \sf{}D = b² - 4 ac⟹D=b
2
−4ac
\begin{gathered} \sf{} \: ( - 2( {a}^{2} - bc {)}^{2} - 4( {c}^{2} - ab)( {b}^{2} - ac) = 0 \\ \\ \sf{}4( {a}^{4} + {b}^{2} {c}^{2} - 2ab {c}^{2} ) - 4( {b}^{2} {c}^{2} - a {c}^{3} - a {b}^{3} + {a}^{2} bc) \\ \\ \sf{}( {a}^{4} + {b}^{2} {c}^{2} - 2ab {c}^{2} ) - ( {b}^{2} {c}^{2} - a {c}^{3} - a {b}^{3} ) + {a}^{2} bc = 0 \\ \\ \sf{} {a}^{4} - 2 {a}^{2} bc + a{c}^{3} + a{b}^{3} - {a}^{2} bc = 0 \\ \\ \sf{} {a}^{4} - 2 {a}^{2} bc + a {c}^{3} + a {b}^{3} - {a}^{2} bc = 0 \\ \\ \sf{}a( {a}^{3} - 2abc + {c}^{3} + {b}^{3} - abc) = 0 \\ \\ \sf{}a( {a}^{3} + {b}^{3} + {c}^{3} - 3abc) = 0 \\ \\ \sf{} \bold{}a = 0 \: \: \: or \: \: \: \: {a}^{3} + {b}^{3} + {c}^{3} = 3abc\end{gathered}
(−2(a
2
−bc)
2
−4(c
2
−ab)(b
2
−ac)=0
4(a
4
+b
2
c
2
−2abc
2
)−4(b
2
c
2
−ac
3
−ab
3
+a
2
bc)
(a
4
+b
2
c
2
−2abc
2
)−(b
2
c
2
−ac
3
−ab
3
)+a
2
bc=0
a
4
−2a
2
bc+ac
3
+ab
3
−a
2
bc=0
a
4
−2a
2
bc+ac
3
+ab
3
−a
2
bc=0
a(a
3
−2abc+c
3
+b
3
−abc)=0
a(a
3
+b
3
+c
3
−3abc)=0
a=0ora
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+b
3
+c
3
=3abc
\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \sf{}Hence \: Proved \:⟹HenceProved