Question

find zero of p(x)=x2+9x+14 and verify believe between zeroes and coefficient ​

Answer 1

Answer:

Zeros of the polynomial are -2 and -7

Explanation:

Given polynomial,

$\implies \: {x}^{2} + 9x + 14 = 0$

On comapring with the general form of equation ax² + bx + c = 0

We get,

• a = 1
• b = 9
• c = 14

By quadratic formula,

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

${ \implies\: x = \dfrac{ - 9 \pm \sqrt{ {(9)}^{2} - 4(1)(14) } }{2(1)} }$

${ \implies\: x = \dfrac{ - 9 \pm \sqrt{ 81 - 56 } }{2} }$

${ \implies\: x = \dfrac{ - 9 \pm \sqrt{25} }{2} }$

${ \implies\: x = \dfrac{ - 9 \pm 5 }{2} }$

${ \implies\: x = \dfrac{ - 9 + 5 }{2} } \: { \rm\: and \: } \: \dfrac{ - 9 - 5}{2}$

${ \implies\: x = \dfrac{ - 4}{2} } \: { \rm\: and \: } \: \dfrac{ - 14}{2}$

${ \implies\: x = - 2 \: { \rm \: and} \: - 7 }$

∴ Zeros of the polynomial are -2 and -7

Now, verification of the zeros and coefficients :-

$\rm \: \bullet \: sum \:of \: zeros : - \\ \implies \: \dfrac{ - b}{a} = - 2 + ( - 7) \\ \implies \: \dfrac{ - 9}{1} = - 2 - 7 \\ \implies \: - 9 = - 9$

And also,

$\rm \: \bullet \: product \: of \: zeros : - \\ \implies \: \dfrac{c}{a} = ( - 2) \ast( - 7)\\ \implies \: \dfrac{14}{1} = 14\\ \implies \:14 = 14$

${\underline{ \underline{ \textsf{ \textbf{hence \: verified}}}}}$

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${{ \rm \bigstar \: General \: form \: of \: a \: quadratic \: equation }= {ax}^{2} + bx + c}$

$\rm \: \bigstar Quadratic \: formula \: : - \\ \implies \: x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$