Question: if {tex}\large \alpha \space and \space \beta{/tex} are the zeros of {tex}\large x^2 + 5 x + 5{/tex} then find the value of: (i) {tex}(\alpha -1)+(\beta -1){/tex},(ii) {tex}\large \frac {\alpha} {\beta} +\frac { \beta} { \alpha}{/tex}
Solution: Roots of the equation {tex}\huge x^2 + 5 x + 5 = {\frac {-5\pm \sqrt 5}{2}}{/tex}
Thus, {tex}\huge \alpha = {\frac {-5+ \sqrt 5}{2}} \space , \space \beta = {\frac {-5- \sqrt 5}{2}}{/tex}
Now, (i) {tex}\huge (\alpha -1)+(\beta -1) \space = \space (\alpha+\beta-2) \space = \space
{\frac {-5+ \sqrt 5}{2}} + {\frac {-5- \sqrt 5}{2}} - 2{/tex}
{tex}\huge \implies
{\frac {-5+ \sqrt 5 -5-\sqrt 5}{2}} - 2{/tex} {tex}\huge \implies
{\frac {-10}{2}} - 2{/tex}
{tex}\huge \implies
{\frac {-10-4}{2}} {/tex} {tex}\huge \implies
{\frac {-14}{2}} {/tex} {tex}\huge \implies\boxed{
-7}{/tex}
Now, (i) {tex}\huge {\frac {\alpha}{\beta}}+{\frac {\beta} {\alpha}} \space = \frac{\alpha ^2 + \beta ^2}{\alpha \beta} {/tex}
{tex}\large \implies
Since, (i)\space The \space sum \space of \space zeroes, \space \alpha+\beta=5 ,(-b/a)\space \space \space ;\space \space \space
(ii)\space The \space product \space of \space zeroes, \space \alpha\beta = 5, \space \space (c/a){/tex}
| {tex}\large \implies
(i) \space\alpha+\beta=5\\ (squaring\space both\space sides\space of\space the\space equation)\\
\implies \alpha^2+2\alpha\beta+\beta^2=25\\
\implies \alpha^2+\beta^2=25-2\alpha\beta\\
(But\space (ii)\space \alpha\beta=5)\\
\therefore\space \alpha^2+\beta^2=25-2\times 5\\
\implies \alpha^2+\beta^2=25-10 \\
\implies \boxed{\alpha^2+\beta^2= 15}{/tex} |
{tex}\huge \implies\therefore
\frac{\alpha^2+\beta^2}{\alpha\beta} \\
\huge \implies \frac{15}{5} \\
\huge \implies \boxed3{/tex} |
Test
Rajan Kumar Pasi 6 years, 7 months ago
Question: if {tex}\large \alpha \space and \space \beta{/tex} are the zeros of {tex}\large x^2 + 5 x + 5{/tex} then find the value of: (i) {tex}(\alpha -1)+(\beta -1){/tex},(ii) {tex}\large \frac {\alpha} {\beta} +\frac { \beta} { \alpha}{/tex}
Solution: Roots of the equation {tex}\huge x^2 + 5 x + 5 = {\frac {-5\pm \sqrt 5}{2}}{/tex}
Thus, {tex}\huge \alpha = {\frac {-5+ \sqrt 5}{2}} \space , \space \beta = {\frac {-5- \sqrt 5}{2}}{/tex}
Now, (i) {tex}\huge (\alpha -1)+(\beta -1) \space = \space (\alpha+\beta-2) \space = \space {\frac {-5+ \sqrt 5}{2}} + {\frac {-5- \sqrt 5}{2}} - 2{/tex}
{tex}\huge \implies {\frac {-5+ \sqrt 5 -5-\sqrt 5}{2}} - 2{/tex} {tex}\huge \implies {\frac {-10}{2}} - 2{/tex}
{tex}\huge \implies {\frac {-10-4}{2}} {/tex} {tex}\huge \implies {\frac {-14}{2}} {/tex} {tex}\huge \implies\boxed{ -7}{/tex}
Now, (i) {tex}\huge {\frac {\alpha}{\beta}}+{\frac {\beta} {\alpha}} \space = \frac{\alpha ^2 + \beta ^2}{\alpha \beta} {/tex}
{tex}\large \implies Since, (i)\space The \space sum \space of \space zeroes, \space \alpha+\beta=5 ,(-b/a)\space \space \space ;\space \space \space (ii)\space The \space product \space of \space zeroes, \space \alpha\beta = 5, \space \space (c/a){/tex}
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