Question

If the zeroes of cubic polynomial x3-15x2+74x-120 are the form a,a+b and a+2b for some positive real number, than the value of a and b are​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{zeroes\;of\;x^3-15x^2+74x-120\;are\;a,\;a+b\;and\;a+2b}$

$\underline{\textbf{To find:}}$

$\textsf{The value of a and b}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,\;\;x^3-15x^2+74x-120}$

$\mathsf{Sum\;of\;zeroes=\dfrac{-(-15)}{1}}$

$\mathsf{a+a+b+a+2b=15}$

$\mathsf{3a+3b=15}$

$\implies\mathsf{a+b=5}$

$\implies\mathsf{b=5-a}$

$\mathsf{Product\;of\;zeroes=120}$

$\implies\mathsf{a(a+b)(a+2b)=120}$

$\implies\mathsf{a(5)(10-a)=120}$

$\implies\mathsf{a(10-a)=24}$

$\implies\mathsf{10a-a^2=24}$

$\implies\mathsf{a^2-10a+24=0}$

$\implies\mathsf{(a-4)(a-6)=0}$

$\implies\mathsf{a=4,6}$

$\mathsf{when\;a=4,}$

$\mathsf{b=5-4}$

$\implies\mathsf{b=1}$

$\mathsf{when\;a=6,}$

$\mathsf{b=5-6}$

$\implies\mathsf{b=-1}$

$\textsf{But a and b are positive real numbers}$

$\therefore\boxed{\mathsf{a=4\;\;and\;\;b=1}}$

$\underline{\textbf{Find more:}}$

Form a cubic polynomial whose zeroes are 2 , 1 , 1.

https://brainly.in/question/16999897

Answer 2

Answer:

Step-by-step explanation:

Given:

${x}^{3} - 15 {x}^{2} + 74x - 120$

Zeroes of cubic polynomial is in the form a,a+b and a+2b .

To find: Find the value of a and b.

Solution:We know that zeroes of cubic polynomial and coefficients of cubic polynomial have some relation.If

$\alpha ,\beta ,\gamma$

are the zeroes of cubic polynomial

$a{x}^{3} + b {x}^{2} + cx + d$

then

$\alpha + \beta + \gamma = \frac{ - b}{a} \\ \\ \alpha \beta + \beta \gamma + \alpha \gamma = \frac{c}{a} \\ \\ \alpha \beta \gamma = \frac{ - d}{a}$

So, here zeros of polynomial are a,a+b and a+2bapply first relation

$a + a + b + a + 2b = 15 \\ \\ or \\ \\ 3a + 3b = 15 \\ \\ or \\ \\ \bold{a + b = 5}...eq1$

From last relation,put value from eq1

$a(a + b)(a + 2b) = 120 \\ \\ a(5)(5 + 5 - a ) = 120 \\ \\ 5a (10 - a) = 120 \\ \\ a (10 - a) = 24 \\ \\ 10a - {a}^{2} - 24 = 0 \\ \\ or \\ \\ {a}^{2} - 10a + 24 = 0 \\ \\ {a}^{2} - 6a - 4a + 24 = 0 \\ \\ a(a - 6) - 4(a - 6) = 0 \\ \\ (a - 6)(a - 4) = 0 \\ \\ either \\ \\ a = 6 \\ \\ or \\ \\ a = 4$

If

$a = 6 \\ \\ then \\ \\ 6 + b = 5 \\ \\ b = - 1$

and if

$a = 4 \\ \\ 4 + b = 5 \\ \\ b = 1$

There are two possibilities of values of a and b

$a = 6 \: b = - 1 \\ \\ a = 4 \: b = 1$

but according to the question values should be positive.

Final answer:a=4 and b= 1

Hope it helps you.

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1)The absolute difference betweenthe integral values of P suchthat(x-p)(x-12)+2 = 0 has integral roots is https://brainly.in/question/40770211