Question

if one zero of the polynomial p(x)=X2+ax+18 is twice the other find the value of a​

Answer 1

Step-by-step explanation:

$let \: \alpha \: and \: \beta \: are \: two \: zeroes \: of \: the \: eqaution \\ \alpha = 2 \beta \\ {x}^{2} + ax + 18 = 0 \\ A = 1, \: \: B = a, \: \: C = 18\\ \color{blue}{ \underline{sum \: of \: zeroes} : } \\ \alpha + \beta = \frac{ - B}{A} \\ 2 \beta + \beta = \frac{ - a}{1} \\ 3 \beta = - a \: - (i) \\ \color{blue}{ \underline{product \: of \: zeroes} : } \\ \alpha \beta = \frac{C}{A} \\ 2 \beta \times \beta = \frac{18}{1} \\ 2 { \beta }^{2} = 18 \\ { \beta }^{2} = 9 \\ \boxed{\beta = ±3} \\ \color{blue}{ \underline{put \: \beta = 3 \: in \: {eq}^{n} \: (i)} : } \\ 3\times 3 = - a \\ \boxed{a = - 9} \\ \color{blue}{ \underline{put \: \beta = -3 \: in \: {eq}^{n} \: (i)} : } \\ \boxed{a = 9} \\ \\ \red{\mathfrak{ \large{\underline{{Hope \: It \: Helps \: You}}}}} \\ \blue{\mathfrak{ \large{\underline{{Mark \: Me \: Brainliest}}}}}$

Answer 2

Given :  one zero of the polynomial p(x)=X2+ax+18 is twice the other

To find  : the value of a

Solution:

Let say Zeroes are α , 2α

Product of Zeroes =  18

=> α * 2α = 18

=> α² = 9

=> α = ± 3

  2α =  ± 6

Sum of zeroes =  α +  2α  = - a

=> a = - 3α

=> a  =  - 3(± 3)

Hence values of a are ± 9

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