if alpha and beta are the zeros of quadratic polynomial p of x²-5x+6 find beta upon alpha plus alpha upon beta and alpha -beta About the author Abigail
[tex] \bf \: Given[/tex] [tex] \sf \to \: {x}^{2} – 5x + 6 = 0[/tex] [tex] \bf \: To \: find [/tex] [tex] \sf \to \: \dfrac{ \beta }{ \alpha } + \dfrac{ \alpha }{ \beta } \: and \: \alpha – \beta [/tex] [tex] \bf \:Now \: Compare \: with [/tex] [tex] \sf \to \: a {x}^{2} + bx + c = 0[/tex] [tex] \bf \: We \: get [/tex] [tex] \sf \to \: a = 1,b = – 5 \: and \: c \: = 6[/tex] [tex] \bf \: We \: know \: that [/tex] [tex] \sf \to \: \alpha + \beta = \dfrac{ – b}{a} = \dfrac{ – ( – 5)}{1} = 5[/tex] [tex] \sf \to \alpha \beta = \dfrac{c}{a} = \dfrac{6}{1} = 6[/tex] [tex] \bf \: Now \: Take [/tex] [tex] \sf \to \: \dfrac{ \beta }{ \alpha } + \dfrac{ \alpha }{ \beta } [/tex] [tex] \bf Now \: Simplify \: the \: equation [/tex] [tex] \sf \to \: \dfrac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } [/tex] [tex]\sf \to \: \dfrac{( \alpha + \beta )^2 – 2 \alpha \beta }{ \alpha \beta }[/tex] [tex] \bf \: Now \: Put \: the \: value [/tex] [tex] \sf \to \: \alpha + \beta = 5 \\ \sf \to \alpha \beta = 6 \: \: \: \: \: \: [/tex] [tex] \bf \: We \: get [/tex] [tex]\sf \to \: \dfrac{25 – 2 \times 6}{6} = \dfrac{25 – 12}{6} = \dfrac{ 13}{6}[/tex] [tex] \bf \: Now \: Take [/tex] [tex] \sf \to \alpha – \beta [/tex] [tex] \bf \: Simplify \: the \: equation [/tex] [tex] \sf \to \sqrt{( \alpha – \beta ) {}^{2} } [/tex] [tex] \sf \to \sqrt{ { \alpha }^{2} + { \beta }^{2} – 2 \alpha \beta } [/tex] [tex] \sf \to \sqrt{( \alpha + \beta ) {}^{2} – 2 \alpha \beta – 2 \alpha \beta } [/tex] [tex] \sf \to \: \sqrt{( \alpha + \beta ) {}^{2} – 4 \alpha \beta } [/tex] [tex] \bf \: Now \: Put \: the \: value [/tex] [tex]\sf \to \: \alpha + \beta = 5 \\ \sf \to \alpha \beta = 6 \: \: \: \: \: \: [/tex] [tex] \sf \to \: \sqrt{(5) { }^{2} – 4 \times 6} [/tex] [tex] \sf \to \sqrt{25 – 24} = \sqrt{1} = 1[/tex] [tex] \bf \:Answer [/tex] [tex]\sf \to \: \dfrac{ \beta }{ \alpha } + \dfrac{ \alpha }{ \beta } = \dfrac{ 13}{6} \: and \: \alpha – \beta = 1[/tex] Reply
Answer: if alpha and beta are the zeros of quadratic polynomial p of x²-5x+6 find beta upon alpha plus alpha upon beta and alpha -beta Reply
[tex] \bf \: Given[/tex]
[tex] \sf \to \: {x}^{2} – 5x + 6 = 0[/tex]
[tex] \bf \: To \: find [/tex]
[tex] \sf \to \: \dfrac{ \beta }{ \alpha } + \dfrac{ \alpha }{ \beta } \: and \: \alpha – \beta [/tex]
[tex] \bf \:Now \: Compare \: with [/tex]
[tex] \sf \to \: a {x}^{2} + bx + c = 0[/tex]
[tex] \bf \: We \: get [/tex]
[tex] \sf \to \: a = 1,b = – 5 \: and \: c \: = 6[/tex]
[tex] \bf \: We \: know \: that [/tex]
[tex] \sf \to \: \alpha + \beta = \dfrac{ – b}{a} = \dfrac{ – ( – 5)}{1} = 5[/tex]
[tex] \sf \to \alpha \beta = \dfrac{c}{a} = \dfrac{6}{1} = 6[/tex]
[tex] \bf \: Now \: Take [/tex]
[tex] \sf \to \: \dfrac{ \beta }{ \alpha } + \dfrac{ \alpha }{ \beta } [/tex]
[tex] \bf Now \: Simplify \: the \: equation [/tex]
[tex] \sf \to \: \dfrac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } [/tex]
[tex]\sf \to \: \dfrac{( \alpha + \beta )^2 – 2 \alpha \beta }{ \alpha \beta }[/tex]
[tex] \bf \: Now \: Put \: the \: value [/tex]
[tex] \sf \to \: \alpha + \beta = 5 \\ \sf \to \alpha \beta = 6 \: \: \: \: \: \: [/tex]
[tex] \bf \: We \: get [/tex]
[tex]\sf \to \: \dfrac{25 – 2 \times 6}{6} = \dfrac{25 – 12}{6} = \dfrac{ 13}{6}[/tex]
[tex] \bf \: Now \: Take [/tex]
[tex] \sf \to \alpha – \beta [/tex]
[tex] \bf \: Simplify \: the \: equation [/tex]
[tex] \sf \to \sqrt{( \alpha – \beta ) {}^{2} } [/tex]
[tex] \sf \to \sqrt{ { \alpha }^{2} + { \beta }^{2} – 2 \alpha \beta } [/tex]
[tex] \sf \to \sqrt{( \alpha + \beta ) {}^{2} – 2 \alpha \beta – 2 \alpha \beta } [/tex]
[tex] \sf \to \: \sqrt{( \alpha + \beta ) {}^{2} – 4 \alpha \beta } [/tex]
[tex] \bf \: Now \: Put \: the \: value [/tex]
[tex]\sf \to \: \alpha + \beta = 5 \\ \sf \to \alpha \beta = 6 \: \: \: \: \: \: [/tex]
[tex] \sf \to \: \sqrt{(5) { }^{2} – 4 \times 6} [/tex]
[tex] \sf \to \sqrt{25 – 24} = \sqrt{1} = 1[/tex]
[tex] \bf \:Answer [/tex]
[tex]\sf \to \: \dfrac{ \beta }{ \alpha } + \dfrac{ \alpha }{ \beta } = \dfrac{ 13}{6} \: and \: \alpha – \beta = 1[/tex]
Answer:
if alpha and beta are the zeros of quadratic polynomial p of x²-5x+6 find beta upon alpha plus alpha upon beta and alpha -beta