Question

6 x square - 11 x + 5.0 sum and product​

Answer 1

Answer:

$sum = \frac{11}{6} \\ \\ product = \frac{5}{6}$

Step-by-step explanation:

$6 {x}^{2} - 11x + 5 = 0 \\ \\ \therefore 6 {x}^{2} - 6x - 5x + 5 = 0 \\ \\ \therefore \: 6x(x - 1) - 5(x - 1) = 0 \\ \\ \therefore \: (6x - 5)(x - 1) = 0$

Take (6x-5)=0

$6x - 5 = 0 \\ \\ \therefore \: 6x = 5 \\ \\ \therefore \: x = \frac{5}{6}$

Take (x-1)=0

$x - 1 = 0 \\ \\ \therefore \: x = 1$

Two Zeros Of Equation Is Given Below

$x = \frac{5}{6} \\ \\ x = 1$

Sum Of Zeros

$= > \frac{5}{6} + 1 \\ \\ = > \frac{5}{6} + \frac{6}{6} \\ \\ = > \frac{11}{6}$

Product Of Zeros

$= > 1 \times \frac{5}{6} \\ \\ = > \frac{5}{6}$

Answer 2

Answer:

$\huge{\boxed{\sf{ \alpha + \beta = \frac{11}{6} (sum \: of \: roots) }}}$

$\huge{\boxed{\sf{ \alpha \beta = \frac{5}{6} (products \: of \: roots)</p><p> }}}$

Step-by-step explanation:

$\huge{\boxed{\sf{SOLUTION-}}}$

$\huge{\boxed{\sf{METHOD(1)}}}$

$6 {x}^{2} - 11x + 5 = 0 \\a = 6 \: \: b = - 11 \: \:a nd \:c = 5 \\ \: we \: have \: to \: find \: sum \\ and \: products \: of \: roots \\ let \: sum \: of \: roots \: be \: \alpha + \beta = \frac{ - b}{a} \\ \alpha + \beta = \frac{ - ( - 11)}{6} \\ \alpha + \beta = \frac{11}{6} (sum \: of \: roots) \\ products \: of \: roots \: be \: \alpha \times \beta = \frac{c}{a} \\ \alpha \times \beta = \frac{5}{6} \\ \alpha \beta = \frac{5}{6} (products \: of \: roots)$

$\huge{\boxed{\sf{METHOD(2)}}}$

$6x^{2}-11x+5=0\\6x^{2}-6x-5x+5=0\\6x(x-1)-5(x-1)=0\\(6x-5)(x-1)=0\\x=\frac{5}{6}\:AND\:1\\sum\:of\:roots=1+\frac{5}{6}\\=)\frac{6+5}{6}\\=)\frac{11}{6}\\product\:of\:roots=1\times\frac{5}{6}\\=)\frac{5}{6}$

$\huge{\boxed{\sf{ \alpha + \beta = \frac{11}{6} (sum \: of \: roots) }}}$

$\huge{\boxed{\sf{ \alpha \beta = \frac{5}{6} (products \: of \: roots)</p><p> }}}$