Question

solve the equation x3 13x2 + 15x + 189 = 0, given that one of the roots exceeds the other by 2.

Answer 1

GIVEN :

The equation $x^3-13x^2+15x+189=0$ , also given that one of the roots exceeds the other by 2.

TO SOLVE :

The given equation $x^3-13x^2+15x+189=0$

SOLUTION :

Given that the equation is $x^3-13x^2+15x+189=0$

Also given that one of the roots exceeds the other by 2.

Let the roots be $\alpha$, $\alpha+2$ and $\beta$

The formula for cubic equation is given  by

$Sum of the roots=\frac{-b}{a}$

We have $Sum of the roots=-\frac{-13}{1}$ here b=-13 and a=1

Sum of the roots$=\alpha+\alpha+2+\beta=-\frac{-13}{1}$

$=13$

$\alpha+(\alpha+2)+\beta=13$

$2\alpha+\beta=13-2$

$\beta=11-2\alpha\hfill (1)$

Now sum of the product of roots taken two at a time,

$\alpha (\alpha+2)+(\alpha+2) \beta+\beta \alpha=\frac{15}{1}$

$\alpha^2+2\alpha+2(\alpha+1) \beta=15\hfill (2)$

The formula for cubic equation is given  by

$Product of the roots=-\frac{c}{a}$

Here a=1 and c=189

$Product of the roots=-\frac{189}{1}$

$Product of the roots=\alpha (\alpha+2)(\beta)=-189$

$\alpha (\alpha+2)(\beta)=-189\hfill (3)$

Now from the equations (1) and (2) we get,

$\alpha^2+2\alpha+2(\alpha+1)(11-2\alpha)=15$  (∵ by (1) $\beta=11-2\alpha$)

$\alpha^2+2\alpha+(2\alpha+2)(11-2\alpha)=15$  

$\alpha^2+2\alpha+22\alpha+22-4\alpha^2-4\alpha=15$

$-3\alpha^2+20\alpha+22-15=0$

$-3\alpha^2+20\alpha+7=0$

$3\alpha^2-20\alpha-7=0$

$3\alpha^2-21\alpha+\alpha-7=0$

$3\alpha (\alpha-7)+1(\alpha-7)=0$

$(3\alpha+1)(\alpha-7)=0$

$(3\alpha+1)=0$ or $(\alpha-7)=0$

$\alpha=-\frac{1}{3}$ or  $\alpha=7$

Now substitute the value of $\alpha=7$ in the equation (1) we get

$\beta=11-2(7)$

$=11-14$

∴ $\beta=-3$

Now the roots $\alpha$, $\alpha+2$ and $\beta$ becomes, $7$, $7+2$ and $-3$

∴ the roots are 7, 9 and -3 of the given cubic equation.

Answer 2

Step-by-step explanation:

GIVEN :

The equation x^3-13x^2+15x+189=0x

3

−13x

2

+15x+189=0 , also given that one of the roots exceeds the other by 2.

TO SOLVE :

The given equation x^3-13x^2+15x+189=0x

3

−13x

2

+15x+189=0

SOLUTION :

Given that the equation is x^3-13x^2+15x+189=0x

3

−13x

2

+15x+189=0

Also given that one of the roots exceeds the other by 2.

Let the roots be \alphaα , \alpha+2α+2 and \betaβ

The formula for cubic equation is given by

Sum of the roots=\frac{-b}{a}Sumoftheroots=

a

−b

We have Sum of the roots=-\frac{-13}{1}Sumoftheroots=−

1

−13

here b=-13 and a=1

Sum of the roots=\alpha+\alpha+2+\beta=-\frac{-13}{1}=α+α+2+β=−

1

−13

=13=13

\alpha+(\alpha+2)+\beta=13α+(α+2)+β=13

2\alpha+\beta=13-22α+β=13−2

\beta=11-2\alpha\hfill (1)β=11−2α\hfill(1)

Now sum of the product of roots taken two at a time,

\alpha (\alpha+2)+(\alpha+2) \beta+\beta \alpha=\frac{15}{1}α(α+2)+(α+2)β+βα=

1

15

\alpha^2+2\alpha+2(\alpha+1) \beta=15\hfill (2)α

2

+2α+2(α+1)β=15\hfill(2)

The formula for cubic equation is given by

Product of the roots=-\frac{c}{a}Productoftheroots=−

a

c

Here a=1 and c=189

Product of the roots=-\frac{189}{1}Productoftheroots=−

1

189

Product of the roots=\alpha (\alpha+2)(\beta)=-189Productoftheroots=α(α+2)(β)=−189

\alpha (\alpha+2)(\beta)=-189\hfill (3)α(α+2)(β)=−189\hfill(3)

Now from the equations (1) and (2) we get,

\alpha^2+2\alpha+2(\alpha+1)(11-2\alpha)=15α

2

+2α+2(α+1)(11−2α)=15 (∵ by (1) \beta=11-2\alphaβ=11−2α )

\alpha^2+2\alpha+(2\alpha+2)(11-2\alpha)=15α

2

+2α+(2α+2)(11−2α)=15

\alpha^2+2\alpha+22\alpha+22-4\alpha^2-4\alpha=15α

2

+2α+22α+22−4α

2

−4α=15

-3\alpha^2+20\alpha+22-15=0−3α

2

+20α+22−15=0

-3\alpha^2+20\alpha+7=0−3α

2

+20α+7=0

3\alpha^2-20\alpha-7=03α

2

−20α−7=0

3\alpha^2-21\alpha+\alpha-7=03α

2

−21α+α−7=0

3\alpha (\alpha-7)+1(\alpha-7)=03α(α−7)+1(α−7)=0

(3\alpha+1)(\alpha-7)=0(3α+1)(α−7)=0

(3\alpha+1)=0(3α+1)=0 or (\alpha-7)=0(α−7)=0

\alpha=-\frac{1}{3}α=−

3

1

or \alpha=7α=7

Now substitute the value of \alpha=7α=7 in the equation (1) we get

\beta=11-2(7)β=11−2(7)

=11-14=11−14

∴ \beta=-3β=−3

Now the roots \alphaα , \alpha+2α+2 and \betaβ becomes, 77 , 7+27+2 and -3−3

∴ the roots are 7, 9 and -3 of the given cubic equation.