Question

Solve the equation x^3-7x^2+14x-8=0 if the roots are in g.P

Answer 1

${x}^{3} - 7 {x}^{2} + 14x - 8 = 0$

• $\frac{a}{r} \: \: \: a \: \: \: \: \: ar \: \: \: \: \\ be \: the \: roots$
• $\frac{a}{r} \times a \times ar = \frac{ -( - 8)}{1}$
• ${a}^{3} = 8$
• a=2
• 2 is a root

Answer 2

The roots of the equation are 1, 2 and 4.

Step-by-step explanation:

The given equation is

$x^3-7x^2+14x-8=0$

It is given that the roots are in G.P.

Let the roots are $\frac{a}{r},a,ar$.

If a polynomial is defined as $P(x)=Ax^3+Bx^2+Cx+D$, then

Sum of roots = -B/A

Product of roots = -D/A

In the given equation A=1,B=-7,C=14 and D=-8.

$\frac{a}{r}\times a\times ar=-\frac{-8}{1}$

$a^3=8$

Taking cube root on both sides.

$a=2$

The value of a is 2.

$\frac{a}{r}+a+ar=-\frac{-7}{1}$

$\frac{2}{r}+2+2r=7$

Multiply both sides by r.

$2+2r+2r^2=7r$

$2+2r-7r+2r^2=0$

$2-5r+2r^2=0$

$2-4r-r+2r^2=0$

$2(1-2r)-r(1-2r)=0$

$(1-2r)(2-r)=0$

Using zero product property we get

$r=\frac{1}{2},2$

If a=2 and r=1/2, then the roots are 4,2 and 1.

If a=2 and r=2, then the roots are 1,2 and 4.

Therefore, the roots of the equation are 1, 2 and 4.

#Learn more

Find the value of k so that the sum of roots of the quadratic equation 3 X square + 2 k + 1 x minus K + 4 equal to zero is equal to the product of the roots.

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