Question

1. Find the zeroes of the quadratic polynomial
6x-13x + 6 and verify the relation between
the zeroes and its coefficients.​

Answer 1

Given :-

6x² - 13x + 6

To Find :-

Zeroes

Solution :-

$\sf :\implies 0 = 6x^2-13x+6$

$\star\sf\underline{Spilt\;the\;middle\;term}$

$\sf :\implies 0 = 6x^2-(9x+4x) + 6$

$\sf :\implies 0 = 6x^2 -9x-4x+6$

$\sf :\implies 0 = 3x(2x - 3)-2(2x - 3)$

$\sf :\implies 0= (2x-3)(3x-2)$

$\sf :\implies x = \dfrac{3}{2}\;\&\; \dfrac{2}{3}$

On comparing the given equation with ax² + bx + c. We get

a = 6

b = -13

c = 6

$\bf \alpha +\beta =\dfrac{-b}{a}$

$\sf :\implies\dfrac{3}{2}+\dfrac{2}{3}=\dfrac{-(-13)}{6}$

$\sf :\implies\dfrac{9 + 4}{6}=\dfrac{13}{6}$

$\sf :\implies \dfrac{13}{6}=\dfrac{13}{6}$

$\bf \alpha \beta =\dfrac{c}{a}$

$\sf :\implies\dfrac{3}{2}\times\dfrac{2}{3}=\dfrac{6}{6}$

$\sf :\implies\dfrac{6}{6}=\dfrac{6}{6}$

$\sf :\implies 1=1$

Hence, Verified

Answer 2

${\boxed{ \LARGE{ \purple{ \rm{ \underline{ Given}}}}}}$

$\bf \Large \hookrightarrow \: \: 6 {x}^{2} \: - \: 13x \: + \: 6$

_________________________

${\boxed{ \LARGE{ \purple{ \rm{ \underline{ To \: \: Find}}}}}}$

Zeroes of the quadratic polynomial and relation between the zeroes and its coefficients.

_______________________________

${\boxed{ \LARGE{ \purple{ \rm{ \underline{ Solution}}}}}}$

$\bf \Large \hookrightarrow \: \: \: 6 {x}^{2} \: - \: 13x \: + \: 6$

$\bf \Large \hookrightarrow \: \:6 {x}^{2} \: - 4x \: - \: 9x \: + \: 6$

$\bf \Large \hookrightarrow \: \:2x \: (3x - 2) \: \: - 3 \: (3x - 2)$

$\bf \Large \hookrightarrow \: \:(2x - 3) \: \: \: (3x - 2)$

$\bf \Large \hookrightarrow \: 2x - 3 = 0 \\ \\\bf \Large \hookrightarrow \: \:x = \: \frac{3}{2}$

$\bf \Large \hookrightarrow \: \:3x - 2 = 0 \\ \\ \bf \Large \hookrightarrow \: \: \: x \: = \: \frac{2}{3}$

$\bf \Large \hookrightarrow \: \: \: x \: = \: \frac{3}{2} \: \: and \: \frac{2}{3}$

Zeroes of the quadratic polynomial 6x² - 13x + 6 is 3/2 and 2/3.

$\:$

$\begin{gathered} {\underline{\boxed{ \bf {\red{ \alpha \: = \: \frac{3}{2} }}}}}\end{gathered} \\ \\ \begin{gathered} {\underline{\boxed{ \bf {\red{ \beta \: = \: \frac{2}{3} }}}}}\end{gathered}$

_______________________

Now , we compare 6x² - 13x + 6 to general equation.

$\:$

☛ General Equation is ax² + bx + c.

$\huge\begin{gathered} {\underline{\boxed{ \sf {\blue{So ,}}}}}\end{gathered}$

⇨a = 6

⇨b = -13

⇨c = 6

________________________

Now , we have to Verify the relationship between Zeros and coefficient.

$\Large \mid \underline {\rm {{{\color{navy}{First \: \: Relation...}}}}} \mid$

$\rm \Large \rightarrow \: \:Sum \: \: of \: \: Zeros$

$\bf \Large \implies \: \:( \alpha + \beta ) \: = \: \frac{ - b}{a}$

$\bf \Large \implies \: \: \bigg( \: \frac{3}{2} \: + \: \frac{2}{3} \: \bigg) \: = \frac{ - 13}{6}$

$\bf \Large \implies \: \: \frac{9 \: + \: 4}{6} \: = \: \frac{ - ( - 13)}{6}$

$\bf \Large \implies \: \: \frac{13}{6} \: = \: \frac{13}{6}$

$\large\begin{gathered} {\underline{\boxed{ \rm {\pink{First \: \: Relation \: \: is \: \: proved}}}}}\end{gathered}$

________________________

$\Large \mid \underline {\rm {{{\color{navy}{Second \: \: Relation...}}}}} \mid$

$\rm \Large \rightarrow \: \:Product \: \: of \: \: Zeros$

$\bf \Large \implies \: \:( \alpha \times \beta ) \: = \: \frac{c}{a}$

$\bf \Large \implies \: \: \frac{3}{2} \: \times \: \frac{2}{3} \: = \: \frac{6}{ 6}$

$\bf \Large \implies \: \: \frac{2 \: \times \: 3}{6} \: = \frac{6}{6}$

$\bf \Large \implies \: \: \: \frac{6}{6} \: = \: \frac{6}{6}$

$\bf \Large \implies \: \: \: \cancel\frac {6} {6} \: = \: \cancel\frac{6} {6}$

$\bf \Large \implies \: \: 1 \: = \: 1$

$\large\begin{gathered} {\underline{\boxed{ \rm {\pink{Second \: \: Relation \: \: is \: \: proved}}}}}\end{gathered}$