`(x+6)/(x^2+9x+18) + (x-p)/(x+7)=0` The equation has real equal roots. Find the possible values of the constant p.

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durbanville | High School Teacher | (Level 2) Educator Emeritus

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The equation has real, equal roots which means `Delta = 0`

First, simplify the equation by factorizing the denominator:

`therefore (x+6)/(x^2+9x+18) + (x-p)/(x+7)=0`  becomes

`(x+6)/((x+6)(x+3))+(x-p)/(x+7)=0`

`therefore 1/(x+3)+(x-p)/(x+7) =0`   The LCD is (x+3)(x+7)

`therefore 1(x+7)+(x+3)(x-p) = 0` (Note: if it was an inequality, we would not drop the denominator)

`therefore x+7+x^2+3x-px-3p=0` Rearrange, including like terms:

`therefore x^2+4x-px+7-3p=0`  

We know that `Delta = b^2-4ac`

so factorize 4x-px as this represents b:

`therefore x^2 +x(4-p)+7-3p=0`

`therefore a =1, b=(4-p) and c= (7-3p)`

Substitute:

`Delta= (4-p)^2-4(1)(7-3p)` (Care with negative signs)

`Delta=16-8p+p^2-28+12p` (Rearrange, add like terms) 

`therefore Delta= p^2+4p-12` Factorize 

`therefore Delta=(p+6)(p-2)`

We know the equation has real, equal roots and that means

`therefore Delta=0` . So

 `therefore (p+6)(p-2)=0`

`therefore p=-6 or p=2`

Ans: p=-6 or p=2 when the equation has real, equal roots.  

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