Question

The two roots of an equation x^3 - 9x^2 + 14x + 24 =0 are in the ratio 3:2 find the roots

Answer 1

it is very clear from the cubic equation f(x)
f(-1)=0,hence -1 is one of the root
as the other roots are in 3:2 ratio ,the roots will be 3a,2a,-1.
sum of the roots =-b/a=9
3a+2a+-1=9
5a=10
a=2
hence the roots are 6,4,-1

Answer 2

Hope u like my process
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Let f(x) =x³ - 9x² +14x +24

Given :-
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The two roots are in the ratio of 3:2

Let common ratio be a

So, the two roots are 3a and 2a respectively.

Now..

f(x) = 0 when x = 3a ; 2a

So..according to the problem
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=> f(3a) = f(2a)

$= > {(3a)}^{3} - 9 {(3a)}^{2} + 14(3a) + 24 = {(2a)}^{3} - 9 {(2a)}^{2} + 14(2a) + 24 \\ \\ or. \: \: 27 {a}^{3} - 81 {a}^{2} + 42a = 8 {a}^{3} - 36 {a}^{2} + 28a \\ \\ or. \: \: 19 {a}^{3} - 45 {a}^{2} + 14 a= 0 \\ \\ or. \: a(19 {a}^{2} - 38a - 7a + 14) = 0 \\ \\ or. \: \: 19a(a - 2) - 7(a - 2) = \frac{0}{a} \\ \\ or. \: \: (19a - 7)(a - 2) = 0 \\ \\ or. \: \: a - 2 = \frac{0}{( {19}a - 7 )} = 0 \\ \\ or. \: \: a = 2$

So.. The required two roots are.,

=> 3a = 3×2 = 6

=> 2a =2×2 = 4

Now to find the last root.
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To know..,

If a and b are two roots of an equation,

Then
(x - a) (x-b) is a multiple of that equation.
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Since the two roots are x = 6,4

So..
(x - 6)(x-4) is a multiple of f(x)
í.e.
(x² - 10x +24)
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$= > {x}^{3} - 9 {x}^{2} + 14x + 24 = 0 \\ \\ or. \: \: {x}^{3} - 10 {x}^{2} + 24x + {x}^{2} - 10x + 24 = 0 \\ \\ or. \: \: \: x( {x}^{2} - 10x + 24) + 1( {x}^{2} - 10x + 24) = 0 \\ \\ or. \: \: (x + 1)( {x}^{2} - 10x + 24) = 0 \\ \\ or. \: \: x + 1 = \frac{0}{( {x}^{2} - 10x + 24) } = 0 \\ \\ or. \: \: x = - 1$
So.. The required third root is - 1

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Thus the required 3 roots are = - 1, 4, 6.

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