[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\sf \: \alpha + \beta = 24 - - - (1) \\ \\ [/tex]
and
[tex]\sf \: \alpha - \beta = 8 - - - (2) \\ \\ [/tex]
On adding equation (1) and (2), we get
[tex]\sf \: 2\alpha = 32 \\ \\ [/tex]
[tex]\bf\implies \:\alpha = 16 \\ \\ [/tex]
On Subtracting equation (2) from (1), we get
[tex]\sf \: 2\beta = 16 \\ \\ [/tex]
[tex]\bf\implies \:\beta = 8 \\ \\ [/tex]
So, the required Quadratic polynomial having zeroes [tex]\alpha [/tex] and [tex]\beta [/tex] is given by
[tex]\sf \: f(x) = {x}^{2} - (\alpha + \beta )x + \alpha \beta \\ \\ [/tex]
On substituting the values of [tex]\alpha [/tex] and [tex]\beta [/tex], we get
[tex]\bf \: f(x) = {x}^{2} - 24x + 128 \\ \\ [/tex]
Verification :-
On comparing with a[tex]x^2 [/tex] + bx + c, we get
[tex]\sf \: a = 1 \\ \\ \sf \: b = - 24 \\ \\ \sf \: c = 128 \\ \\ [/tex]
Now,
[tex]\sf \: Sum \: of \: zeroes, \: \alpha + \beta = 16 + 8 = 24 = \dfrac{ - ( - 24)}{1} = - \frac{b}{a} \\ \\ [/tex]
Hence,
[tex]\rm\implies \:\boxed{{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}} \\ \\ [/tex]
[tex]\sf \:Product \: of \: zeroes, \: \alpha \beta = 16 \times 8 = 128= \dfrac{128}{1} = \frac{c}{a} \\ \\ [/tex]
[tex]\rm\implies \:\boxed{{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}} \\ \\ [/tex]
Hence, Verified
[tex]\rule{190pt}{2pt}[/tex]
[tex] {{ \mathfrak{Additional\:Information}}}[/tex]
[tex]{\sf :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then} \\ \\ [/tex]
[tex]\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}} \\ \\ [/tex]
[tex]\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}} \\ \\ [/tex]
[tex]\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}} \\ \\ [/tex]
Answer:
Given :
[tex]\small \mapsto \sf \alpha\: and\: \beta\: are\: the\: zeroes\: of\: the\: quadratic\: polynomial\: .\\[/tex]
Also Given :
[tex]\leadsto \bf \alpha + \beta =\: 24\: ------\: (Equation\: No\: 1)\\[/tex]
[tex]\leadsto \bf \alpha - \beta =\: 8\: ------\: (Equation\: No\: 2)\\[/tex]
By adding the equation no 1 with the equation no 2 we get,
[tex]\implies \sf \alpha + \beta + (\alpha - \beta) =\: 24 + 8\\[/tex]
[tex]\implies \sf \alpha + \beta + \alpha - \beta =\: 32\\[/tex]
[tex]\implies \sf \alpha + \alpha {\cancel{+ \beta}} {\cancel{- \beta}} =\: 32\\[/tex]
[tex]\implies \sf \alpha + \alpha =\: 32[/tex]
[tex]\implies \sf 2\alpha =\: 32[/tex]
[tex]\implies \sf \alpha =\: \dfrac{32}{2}[/tex]
[tex]\implies \sf\bold{\alpha =\: 16}[/tex]
Now, by putting the value of α = 16 in the equation no 1 we get,
[tex]\implies \sf \alpha + \beta =\: 24[/tex]
[tex]\implies \sf 16 + \beta =\: 24[/tex]
[tex]\implies \sf \beta =\: 24 - 16[/tex]
[tex]\implies \sf\bold{\beta =\: 8}[/tex]
Hence, the value of α is 16 and the value of β is 8 .
So, the sum of roots and product of roots will be :
❒ In case of Sum Of Roots :
[tex]\mapsto \sf \alpha + \beta[/tex]
[tex]\mapsto \sf 16 + 8[/tex]
[tex]\mapsto \bf 24[/tex]
❒ In case of Product Of Roots :
[tex]\mapsto \sf \alpha\beta[/tex]
[tex]\mapsto \sf (16)(8)[/tex]
[tex]\mapsto \sf 16 \times 8[/tex]
[tex]\mapsto \bf 128[/tex]
Now, we have to find the quadratic polynomial :
According to the question by using the formula we get,
[tex]\small \implies \sf\boxed{\bold{x^2 - (Sum\: of\: Roots)x + Product\: of\: Roots}}\\[/tex]
[tex]\implies \sf x^2 - (\alpha + \beta)x + (\alpha\beta)\\[/tex]
[tex]\implies \sf x^2 - (24)x + (128)[/tex]
[tex]\implies \sf\bold{\underline{x^2 - 24x + 128}}[/tex]
[tex]\therefore[/tex] The required quadratic polynomial is x² - 24x + 128 .
Now, we have to verify the zeroes between the zeroes and co-efficient of the polynomial :
Given Equation :
[tex]\longrightarrow \bf x^2 - 24x + 128[/tex]
By comparing with ax² + bx + c we get,
where,
❒ In case of Sum of Roots :
[tex]\implies \sf\boxed{\bold{Sum\: of\: Roots\: (\alpha + \beta) =\: \dfrac{- b}{a}}}\\[/tex]
So, by putting those values we get,
[tex]\implies \sf 24 =\: \dfrac{- (- 24)}{1}[/tex]
[tex]\implies \sf 24 =\: \dfrac{24}{1}[/tex]
[tex]\implies \sf\bold{24 =\: 24}[/tex]
[tex]\implies \sf L.H.S. =\: R.H.S[/tex]
Hence, Verified !!
❒ In case of Product of Roots :
[tex]\implies \sf\boxed{\bold{Product\: of\: Roots\: (\alpha\beta) =\: \dfrac{c}{a}}}\\[/tex]
[tex]\implies \sf 128 =\: \dfrac{128}{1}[/tex]
[tex]\implies \sf\bold{128 =\: 128}[/tex]
[tex]\implies \sf L.H.S. =\: R.H.S.[/tex]
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Answers & Comments
Verified answer
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
[tex]\sf \: \alpha + \beta = 24 - - - (1) \\ \\ [/tex]
and
[tex]\sf \: \alpha - \beta = 8 - - - (2) \\ \\ [/tex]
On adding equation (1) and (2), we get
[tex]\sf \: 2\alpha = 32 \\ \\ [/tex]
[tex]\bf\implies \:\alpha = 16 \\ \\ [/tex]
On Subtracting equation (2) from (1), we get
[tex]\sf \: 2\beta = 16 \\ \\ [/tex]
[tex]\bf\implies \:\beta = 8 \\ \\ [/tex]
So, the required Quadratic polynomial having zeroes [tex]\alpha [/tex] and [tex]\beta [/tex] is given by
[tex]\sf \: f(x) = {x}^{2} - (\alpha + \beta )x + \alpha \beta \\ \\ [/tex]
On substituting the values of [tex]\alpha [/tex] and [tex]\beta [/tex], we get
[tex]\bf \: f(x) = {x}^{2} - 24x + 128 \\ \\ [/tex]
Verification :-
On comparing with a[tex]x^2 [/tex] + bx + c, we get
[tex]\sf \: a = 1 \\ \\ \sf \: b = - 24 \\ \\ \sf \: c = 128 \\ \\ [/tex]
Now,
[tex]\sf \: Sum \: of \: zeroes, \: \alpha + \beta = 16 + 8 = 24 = \dfrac{ - ( - 24)}{1} = - \frac{b}{a} \\ \\ [/tex]
Hence,
[tex]\rm\implies \:\boxed{{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}} \\ \\ [/tex]
Now,
[tex]\sf \:Product \: of \: zeroes, \: \alpha \beta = 16 \times 8 = 128= \dfrac{128}{1} = \frac{c}{a} \\ \\ [/tex]
[tex]\rm\implies \:\boxed{{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}} \\ \\ [/tex]
Hence, Verified
[tex]\rule{190pt}{2pt}[/tex]
[tex] {{ \mathfrak{Additional\:Information}}}[/tex]
[tex]{\sf :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then} \\ \\ [/tex]
[tex]\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}} \\ \\ [/tex]
[tex]\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}} \\ \\ [/tex]
[tex]\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}} \\ \\ [/tex]
Answer:
Given :-
To Find :-
Solution :-
Given :
[tex]\small \mapsto \sf \alpha\: and\: \beta\: are\: the\: zeroes\: of\: the\: quadratic\: polynomial\: .\\[/tex]
Also Given :
[tex]\leadsto \bf \alpha + \beta =\: 24\: ------\: (Equation\: No\: 1)\\[/tex]
[tex]\leadsto \bf \alpha - \beta =\: 8\: ------\: (Equation\: No\: 2)\\[/tex]
By adding the equation no 1 with the equation no 2 we get,
[tex]\implies \sf \alpha + \beta + (\alpha - \beta) =\: 24 + 8\\[/tex]
[tex]\implies \sf \alpha + \beta + \alpha - \beta =\: 32\\[/tex]
[tex]\implies \sf \alpha + \alpha {\cancel{+ \beta}} {\cancel{- \beta}} =\: 32\\[/tex]
[tex]\implies \sf \alpha + \alpha =\: 32[/tex]
[tex]\implies \sf 2\alpha =\: 32[/tex]
[tex]\implies \sf \alpha =\: \dfrac{32}{2}[/tex]
[tex]\implies \sf\bold{\alpha =\: 16}[/tex]
Now, by putting the value of α = 16 in the equation no 1 we get,
[tex]\implies \sf \alpha + \beta =\: 24[/tex]
[tex]\implies \sf 16 + \beta =\: 24[/tex]
[tex]\implies \sf \beta =\: 24 - 16[/tex]
[tex]\implies \sf\bold{\beta =\: 8}[/tex]
Hence, the value of α is 16 and the value of β is 8 .
So, the sum of roots and product of roots will be :
❒ In case of Sum Of Roots :
[tex]\mapsto \sf \alpha + \beta[/tex]
[tex]\mapsto \sf 16 + 8[/tex]
[tex]\mapsto \bf 24[/tex]
❒ In case of Product Of Roots :
[tex]\mapsto \sf \alpha\beta[/tex]
[tex]\mapsto \sf (16)(8)[/tex]
[tex]\mapsto \sf 16 \times 8[/tex]
[tex]\mapsto \bf 128[/tex]
Now, we have to find the quadratic polynomial :
Given :
According to the question by using the formula we get,
[tex]\small \implies \sf\boxed{\bold{x^2 - (Sum\: of\: Roots)x + Product\: of\: Roots}}\\[/tex]
[tex]\implies \sf x^2 - (\alpha + \beta)x + (\alpha\beta)\\[/tex]
[tex]\implies \sf x^2 - (24)x + (128)[/tex]
[tex]\implies \sf\bold{\underline{x^2 - 24x + 128}}[/tex]
[tex]\therefore[/tex] The required quadratic polynomial is x² - 24x + 128 .
Now, we have to verify the zeroes between the zeroes and co-efficient of the polynomial :
Given Equation :
[tex]\longrightarrow \bf x^2 - 24x + 128[/tex]
By comparing with ax² + bx + c we get,
where,
❒ In case of Sum of Roots :
[tex]\implies \sf\boxed{\bold{Sum\: of\: Roots\: (\alpha + \beta) =\: \dfrac{- b}{a}}}\\[/tex]
So, by putting those values we get,
[tex]\implies \sf 24 =\: \dfrac{- (- 24)}{1}[/tex]
[tex]\implies \sf 24 =\: \dfrac{24}{1}[/tex]
[tex]\implies \sf\bold{24 =\: 24}[/tex]
[tex]\implies \sf L.H.S. =\: R.H.S[/tex]
Hence, Verified !!
❒ In case of Product of Roots :
[tex]\implies \sf\boxed{\bold{Product\: of\: Roots\: (\alpha\beta) =\: \dfrac{c}{a}}}\\[/tex]
So, by putting those values we get,
[tex]\implies \sf 128 =\: \dfrac{128}{1}[/tex]
[tex]\implies \sf\bold{128 =\: 128}[/tex]
[tex]\implies \sf L.H.S. =\: R.H.S.[/tex]
Hence, Verified !!