Question

find the zeroes of the polynomial x square -5x+6 and verify the relationship between the zeros and the coefficients​

Answer 1

Answer:

Zeros of the polynomial is 3 and 2

Explanation:

Given polynomial,

$\rm \implies \: {x}^{2} - 5x + 6$

On comparing with the general form of a quadratic equation ax² + bx + c

We get,

• a = 1
• b = -5
• c = 6

By quadratic formula,

$\rm \implies \: x = \dfrac{ - b \pm \sqrt{ {(b)}^{2} - 4ac} }{2a}$

$\rm \implies \: x = \dfrac{ - ( - 5) \pm \sqrt{ {( - 5)}^{2} - 4(1)(6)} }{2(1)}$

$\rm \implies \: x = \dfrac{ 5 \pm \sqrt{ 25- 24} }{2}$

$\rm \implies \: x = \dfrac{ 5 \pm \sqrt{1} }{2}$

$\rm \implies \: x = \dfrac{ 5 + 1 }{2} \: and \: \dfrac{5 - 1}{2}$

$\rm \implies \: x = \dfrac{6}{2} \: and \: \dfrac{4}{2}$

$\rm \therefore \: x = 3 \: and \: 2$

∴ Zeros of the polynomial are 3 and 2

Verification,

$\rm \bullet\: sum \: of \: zeros = \dfrac{ - b}{a}$

$\rm \implies \: 3 + 2 = \dfrac{ - ( - 5)}{1}$

$\rm \implies \: 5= \dfrac{ 5}{1}$

$\rm \implies \: 5= { 5}$

And also,

$\rm \bullet \: product \: of \: zeros = \dfrac{c}{a}$

$\rm \implies \: 3 \ast 2 = \dfrac{6}{1}$

$\rm \implies \: 6 = 6$

Hence, verified !!

Answer 2

$\huge \blue \diamond \: \huge \underbrace {\textrm{{{\color{brown}{Given}}}}} \: \blue\diamond$

We have given Polynomial

$\bf \Large \hookrightarrow \: \: {x}^{2} \: - \: 5x \: + \: 6$

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$\huge \blue \diamond \: \huge \underbrace {\textrm{{{\color{brown}{To \: Find}}}}} \: \blue\diamond$

Zeroes of the polynomial and we have to verify the relationship between the zeros and the coefficients.

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$\huge \blue \diamond \: \huge \underbrace {\textrm{{{\color{brown}{Solution}}}}} \: \blue\diamond$

Now , we using factorisation method to find the Zeros of polynomial x² - 5x + 6.

$\bf \large \rightarrow \: \: {x}^{2} \: - \: 5x \: + \: 6$

By Splitting the middle term :

$\bf \large \rightarrow \: \: {x}^{2} \: - \: 5x \: + \: 6$

$\bf \large \rightarrow \: \: {x}^{2} \: - \: 2x \: - \: 3x \: + \: 6$

$\bf \large \rightarrow \: \: x \: (x - 2) \: \: \: - 3 \: (x - 2)$

$\bf \large \rightarrow \: \: ( \: x \: - \: 3 \: ) \: \: \: \: \: \: \: ( \: x \: - \: 2 \: )$

$\bf \large \rightarrow \: x \: - \: 3 \: = \: 0 \: \: \: \: ; \: \: \: x \: - \: 2 \: = \: 0$

$\bf \large \rightarrow \: \: x \: = \: 3 \: \: \: \: \: \: ; \: \: \: \: \: \: x \: = \: 2$

$\bf \large \rightarrow \: \: x \: = \: 3 \: \: and \: \: 2$

Zeros of the polynomial is 3 and 2 .

$\Large\begin{gathered} {\underline{\boxed{ \bf{\red{ \alpha \: = \: 3}}}}}\end{gathered}$

$\Large\begin{gathered} {\underline{\boxed{ \bf {\red{ \beta \: = \: 2}}}}}\end{gathered}$

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Now , comparing x² - 5x + 6 to the general form of a quadratic equation.

$\:$

General form of quadratic equation - ax² + bx + c.

Then ,

$\bf \large \implies \: \: a \: \: = \: \: 1$

$\bf \large \implies \: \:b \: \: = \: \: - 5$

$\bf \large \implies \: \:c \: \: = \: \: 6$

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Now we have to verify the relationship between the zeros and the coefficients.

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

$\bf \large \implies \: \:Sum \: \: of \: \: Zeros$

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

$\bf \Large \: \:( \: 3 \: + \: 2 \: ) \: = \: \frac{ - \: ( - 5)}{1}$

$\bf \Large \: \: \implies \: \: \: 5 \: = \: 5$

${\boxed{ \Large{ \blue{ \bf{ \underline{ LHS = RHS}}}}}}$

First Relation is proved.

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

$\bf \large \implies \: \:Product \: \: of \: \: Zeros$

$\bf \Large \: \: \alpha \: \beta \: = \: \frac{c}{a}$

$\bf \Large \: \:3 \: \times \:2 \: = \: \frac{6}{1}$

$\bf \Large \: \: \implies \: \: \: 6 \: = \: 6$

${\boxed{ \Large{ \blue{ \bf{ \underline{ LHS = RHS}}}}}}$

Second Relation is proved.

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$\huge \blue \diamond \: \huge \underbrace {\textrm{{{\color{brown}{Note}}}}} \: \blue\diamond$

You can solve this question by second method ( Quadratic formula ) also with is already answered by @taslimuddinahmed60.