Answers
Answer:
(i) (b² - 2ac)/c²(ii) (3abc - b³)/a³Step-by-step explanation:
α and β are the roots of the equation:
ax² + bx + c = 0Sum of the roots is:
α + b = -b/aProduct of the roots is:
αβ = c/aSolving the following expressions:
(i)
1/α² + 1/β² =(α² + β²) / α²β² =((α + β)² - 2αβ) / (αβ)² = ((-b/a)² - 2c/a) / (c/a)² = (b²/a² - 2c/a) * a²/c² = b²/c² - 2ac/c² =(b² - 2ac)/c²----------------
(ii)
α³ + β³ =(α + β)(α² - αβ + β²) =(α + β)((α + β)² - 3αβ) = (α + β)³ - 3αβ(α + β) =(-b/a)³ - 3(c/a)(-b/a) =-b³/a³ + 3bc/a²= 3abc/a³ - b³/a³=(3abc - b³)/a³[tex] \huge \underline{\tt Question} :[/tex]
If α and β are the roots of the equation ax² + bx + c = 0, where a, b and c are constants such that a ≠ 0, find in terms of a, b and c expressions for :
[tex] \tt \dfrac{1}{\alpha ^2} + \dfrac{1}{\beta ^2}[/tex]α³ + β³[tex] \\ [/tex]
[tex] \huge \underline{\tt Answer} :[/tex]
[tex] \bf \dfrac{1}{\alpha ^2} + \dfrac{1}{\beta ^2} = \dfrac{b^2 - 2ac}{c^2 }[/tex][tex] \bf \alpha ^3 + \beta ^3 = \dfrac{ - b^3 + 3abc}{a^3}[/tex][tex] \\ [/tex]
[tex] \huge \underline{\tt Explanation} :[/tex]
As, α and β are the roots of the equation ax² + bx + c = 0
We know that :
[tex] \underline{\boxed{\bf{Sum \: of \: roots = \dfrac{- coefficient \: of \: x}{coefficient \: of \: x^2}}}}[/tex][tex] \underline{\boxed{\bf{Product \: of \: roots = \dfrac{constant \: term}{coefficient \: of \: x^2}}}}[/tex][tex] \tt : \implies \alpha + \beta = \dfrac{-b}{a}[/tex]
and
[tex] \tt : \implies \alpha\beta = \dfrac{c}{a}[/tex]
[tex] \\ [/tex]
Now, let's solve given values :
[tex] \bf \: \: \: \: 1. \: \dfrac{1}{\alpha ^2} + \dfrac{1}{\beta ^2}[/tex]
[tex] \tt : \implies \dfrac{\beta ^2 + \alpha ^2}{\alpha ^2 \beta ^2}[/tex]
[tex] \tt : \implies \dfrac{\alpha ^2 + \beta ^2}{\alpha ^2 \beta ^2}[/tex]
[tex] \\ [/tex]
Now, by using identity :
[tex] \underline{\boxed{\bf{a^2+ b^2 = (a+b)^2 -2ab}}}[/tex][tex] \tt : \implies \dfrac{(\alpha + \beta)^2 - 2 \alpha\beta}{(\alpha\beta)^2}[/tex]
[tex] \\ [/tex]
Now, by substituting values of :
[tex] \underline{\boxed{\bf{\alpha + \beta = \dfrac{-b}{a}}}}[/tex][tex] \underline{\boxed{\bf{\alpha\beta = \dfrac{c}{a}}}}[/tex][tex] \tt : \implies \dfrac{\Bigg(\dfrac{-b}{a}\Bigg)^2 - 2 \times \dfrac{c}{a}}{\Bigg(\dfrac{c}{a}\Bigg)^2}[/tex]
[tex] \tt : \implies \dfrac{\dfrac{b^2}{a^2} - \dfrac{2c}{a}}{\dfrac{c^2}{a^2}}[/tex]
[tex]\tt : \implies \dfrac{\dfrac{b^2}{a^2} - \dfrac{2ac}{a^{2} }}{\dfrac{c^2}{a^2}}[/tex]
[tex]\tt : \implies \dfrac{\dfrac{b^2 - 2ac}{a^2 }}{\dfrac{c^2}{a^2}}[/tex]
[tex]\tt : \implies \dfrac{b^2 - 2ac}{\cancel{a^2} } \times \dfrac{ \cancel{a^2}}{c^2}[/tex]
[tex]\tt : \implies \dfrac{b^2 - 2ac}{c^2 }[/tex]
[tex] \\ [/tex]
[tex] \underline{\bf Hence, \: \dfrac{1}{\alpha ^2} + \dfrac{1}{\beta ^2} = \dfrac{b^2 - 2ac}{c^2 }}[/tex]
[tex] \\ [/tex]
[tex] \bf \: \: \: \: 2. \: \alpha ^3 + \beta ^3 [/tex]
[tex] \\ [/tex]
By using identity :
[tex] \underline{\boxed{\bf{a^3+ b^3 = (a+b)(a^2 -ab + b^2)}}}[/tex][tex] \tt : \implies (\alpha + \beta)(\alpha ^2 - \alpha\beta + \beta ^2)[/tex]
[tex] \tt : \implies (\alpha + \beta)(\alpha ^2 + \beta ^2 - \alpha\beta)[/tex]
[tex] \\ [/tex]
By using identity :
[tex] \underline{\boxed{\bf{a^2+ b^2 = (a+b)^2 -2ab}}}[/tex][tex] \tt : \implies (\alpha + \beta)(\alpha + \beta)^2 -2 \alpha\beta - \alpha\beta)[/tex]
[tex] \tt : \implies (\alpha + \beta)((\alpha + \beta)^2 -3 \alpha\beta)[/tex]
[tex] \\ [/tex]
Now, by substituting values of :
[tex] \underline{\boxed{\bf{\alpha + \beta = \dfrac{-b}{a}}}}[/tex][tex] \underline{\boxed{\bf{\alpha\beta = \dfrac{c}{a}}}}[/tex][tex] \tt : \implies \Bigg(\dfrac{-b}{a}\Bigg)\Bigg( \bigg(\dfrac{-b}{a} \bigg)^2 -3 \times \dfrac{c}{a}\Bigg)[/tex]
[tex] \tt : \implies \Bigg(\dfrac{-b}{a}\Bigg)\Bigg(\dfrac{b^2}{a^2} - \dfrac{3c}{a}\Bigg)[/tex]
[tex]\tt : \implies \Bigg(\dfrac{-b}{a}\Bigg)\Bigg(\dfrac{b^2}{a^2} - \dfrac{3ac}{a^{2} }\Bigg)[/tex]
[tex]\tt : \implies \Bigg(\dfrac{-b}{a}\Bigg)\Bigg(\dfrac{b^2 - 3ac}{a^2} \Bigg)[/tex]
[tex]\tt : \implies \dfrac{-b}{a} \times \dfrac{b^2 - 3ac}{a^2} [/tex]
[tex]\tt : \implies \dfrac{ - b(b^2 - 3ac)}{a \times a^2} [/tex]
[tex]\tt : \implies \dfrac{ - b^3 + 3abc}{a^3} [/tex]
[tex] \\ [/tex]
[tex] \underline{\bf Hence, \: \alpha ^3 + \beta ^3 = \dfrac{ - b^3 + 3abc}{a^3}}[/tex]
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Answers
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Step-by-step explanation:
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Answers
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Answers
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Answers
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Calculating the value of x:When considering unknown values, the value of x is used. In algebra, the letter "x" is frequently used to denote an unknown value. It's called a "variable" or an "unknown" in other circumstances.[tex]\to[/tex] 2x=17
[tex]\to[/tex] x=[tex]\frac{17}{2}[/tex]
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brainly.com/question/14170328
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Answers
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Answers
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Step-by-step explanation:
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Answers
Answer:
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Step-by-step explanation:
there
Answer:
[tex]Evelyn \: knit \: made \to \\ \boxed{12 }\: hats \: and \: \boxed{8 } \: scarves[/tex]
Step-by-step explanation:
[tex]let \: the \: hats \: be \to \: h \\ let \: the \: scarves \: be \to \: s \\ then \to \\ 5h + 15s = 180.....(eq .1) \\ 110h + 150s = 2,520.....(eq .2) \\ from \: .....(eq .1) \\ 15s = 180 - 5h \\ s = \frac{180 - 5h}{15} .....(eq .3) \\ equating \: .....(eq .3) \: in \: .....(eq .2) \\ 110h + 150( \frac{180 - 5h}{15} ) = 2,520 \\ 110h + 10(180 - 5h) = 2,520 \\ 110h + 1800 - 50h= 2,520 \\ 60h = 2520 - 1800 \\ 60h = 720 \\ \boxed{h = 12} \\ applying \: \boxed{h} \: in \: .....(eq .3) \\ s = \frac{180 - 5(h)}{15} \\ \\ s = \frac{180 - 5(12)}{15} \\ \\ s = \frac{120}{15} \\ \boxed{ s = 8}[/tex]
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