How do you find the value of alpha minus beta?
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alpha and beta are the roots of x square + kx + 12 is equal to zero and Alpha minus beta is equal to 1 by 2 find the value of k it into the solution compare the equation with the general form of quadratic equation X square + bx + c is equals to zero we get a is equals to 1 b is equals to K C is equals to 12 OK so the sum of the roots alpha + beta is given by mines bi Baje which is equals to minus ke the product of roots alpha beta is given by sea way which is equals to 12 and in the question whether given that of minus beta is equals to 1 Alpha minus beta is equals to 1 ok we can write alpha + beta whole square is equals to Show
Alpha minus beta whole square + 4 alpha beta OK then forget alpha + beta minus ke whole square is equals to minus b square + 4 into Pal divided ke score is equals to 49 can we get the value of k plus or minus 7 Identity 1: α2 + β2$$ \boxed{ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \phantom{.} } $$ Deriving the formula: \begin{align*} \text{Since } (a + b)^2 & = a^2 + 2ab + b^2, \\ \\ (\alpha + \beta)^2 & = \alpha^2 + 2\alpha\beta + \beta^2 \\ (\alpha + \beta)^2 - 2\alpha \beta & = \alpha^2 + \beta^2 \\ \\ \therefore \alpha^2 + \beta^2 & = (\alpha + \beta)^2 - 2\alpha \beta \end{align*} Identity 2: (α - β)2$$ \boxed{ (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \phantom{.} } $$ Deriving the formula: \begin{align*} \text{Since } (a - b)^2 & = a^2 - 2ab + b^2, \\ \\ (\alpha - \beta)^2 & = \alpha^2 - 2\alpha\beta + \beta^2 \\ & = \underbrace{\alpha^2 + \beta^2}_\text{Identity 1} - 2\alpha \beta \\ & = [(\alpha + \beta)^2 - 2 \alpha\beta] - 2\alpha\beta \\ & = (\alpha + \beta)^2 - 4 \alpha\beta \\ \\ \therefore (\alpha - \beta)^2 & = (\alpha + \beta)^2 - 4\alpha \beta \end{align*}
Identity 3: α3 + β3Two versions: $$ \boxed{ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \phantom{.} } $$ $$ \boxed{ \alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 3\alpha\beta] \phantom{.} } $$ Converting from the first version to the second version: \begin{align*} \alpha^3 + \beta^3 & = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \\ & = (\alpha + \beta) ( \underbrace{ \alpha^2 + \beta^2 }_\text{Identity 1} - \alpha \beta ) \\ & = (\alpha + \beta) [ (\alpha + \beta)^2 - 2 \alpha \beta - \alpha \beta] \\ & = (\alpha + \beta) [ (\alpha + \beta)^2 - 3 \alpha \beta] \end{align*} Identity 4: α3 - β3Two versions: $$ \boxed{ \alpha^3 - \beta^3 = (\alpha - \beta)(\alpha^2 + \alpha\beta + \beta^2) \phantom{.} } $$ $$ \boxed{ \alpha^3 + \beta^3 = (\alpha - \beta)[(\alpha + \beta)^2 - \alpha\beta] \phantom{.} } $$ Converting from the first version to the second version: \begin{align*} \alpha^3 - \beta^3 & = (\alpha - \beta)(\alpha^2 + \alpha\beta + \beta^2) \\ & = (\alpha - \beta)( \underbrace{\alpha^2 + \beta^2}_\text{Identity 1} + \alpha \beta) \\ & = (\alpha - \beta) [(\alpha + \beta)^2 - 2\alpha\beta + \alpha\beta ] \\ & = (\alpha - \beta) [(\alpha + \beta)^2 - \alpha\beta ] \end{align*} Identity 5: α4 + β4$$ \boxed{ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2 \phantom{.} } $$ Deriving the formula using identity 1: \begin{align*} \text{Identity 1: } \alpha^2 + \beta^2 & = (\alpha + \beta)^2 - 2\alpha \beta \\ \\ \text{Replace } \alpha \text{ by } \alpha^2 & \text{ and } \beta \text{ by } \beta^2, \\ \\ (\alpha^2)^2 + (\beta^2)^2 & = (\alpha^2 + \beta^2)^2 - 2(\alpha^2)(\beta^2) \\ \alpha^4 + \beta^4 & = (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2 \end{align*} Identity 6: α4 - β4$$ \boxed{ \alpha^4 - \beta^4 = (\alpha^2 + \beta^2)(\alpha + \beta)(\alpha - \beta) \phantom{.} } $$ Deriving the formula using the identity $a^2 - b^2 = (a + b)(a - b)$: \begin{align*} \alpha^4 - \beta^4 & = (\alpha^2)^2 - (\beta^2)^2 \\ & = (\alpha^2 + \beta^2) (\alpha^2 - \beta^2) \\ & = (\alpha^2 + \beta^2) (\alpha + \beta) (\alpha - \beta) \end{align*}
For the equation #x^2+lx+m=0# sum of roots is #-l# and product of roots is #m#. Therefore, as for #x^2-22x+105=0# roots are #alpha# and #beta# hence #alpha+beta=-(-22)=22# and #alphabeta=105# As #(alpha+beta)^2=(alpha-beta)^2+4alphabeta# #22^2=(alpha-beta)^2+4*105# or #(alpha-beta)^2=22^2-420=484-420=64# and #alpha-beta=8# One could say that we can also have #alpha-beta=-8#, but observe that #alpha# and #beta# are not in any particular order. The roots of equation are #15# and#7# and their #alpha-beta# could be #15-7# as well as #7-15#, it deends on what you choose as #alpha# and #beta#. If #(alpha>beta)#, then ,#(alpha-beta)=8# If the quadratic equation #ax^2+bx+c=0#, has roots #alpha and beta, #then #alpha+beta=-b/a and alpha*beta=c/a.# #x^2-22x+105=0=>a=1 , b=-22 , c=105# So, #(alpha-beta)=sqrt((22)^2-4(105))# #(alpha-beta)=sqrt(484-420)=sqrt64=8# What is the formula of finding alpha minus beta?⇒ α−β=a±
What is alpha minus beta square?So, (α - β)² = (α + β)² - 4αβ
What is the value of one by alpha minus 1 by Beta?Hence, 1α−1β=−32 or 32.
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