Sum of reciprocals.

Level pending

For the equation of the form: a x 2 + b x + c = 0 ax^2+bx+c=0 ,determine

1 x 1 + 1 x 2 \frac{1}{x_1}+\frac{1}{x_2} where x 1 x_1 and x 2 x_2 are the roots if a = 5 a=5 , b = 3 b=3 and c = − 1 c=-1 .


The answer is 3.

Lorenc Bushi by Lorenc Bushi , 21, Albania

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1 solution

Prasun Biswas
Feb 3, 2014

We shall here use the quadratic formula to find the roots x 1 x_{1} and x 2 x_{2} .

According to the formula, if a quadratic equation is in the form a x 2 + b x + c = 0 ax^{2}+bx+c=0 and has its discriminant ( D ) = b 2 − 4 a c ≥ 0 (D)=b^{2}-4ac \geq 0 , then the equation has real roots and the roots are = − b ± D 2 a =\large{\frac{-b\pm \sqrt{D}}{2a}} .

Here, it is said that the equation is in a x 2 + b x + c = 0 ax^{2}+bx+c=0 form with a = 5 , b = 3 , c = ( − 1 ) a=5,b=3,c=(-1) , so discriminant ( D ) = 3 2 − 4 × 5 × ( − 1 ) = 9 + 20 = 29 ≥ 0 (D)=3^{2}-4\times 5\times (-1)=9+20=29 \geq 0 , so the equation has real roots.

So, the roots are = − 3 ± 29 2 × 5 = − 3 ± 29 10 =\large{\frac{-3\pm \sqrt{29}}{2\times 5} = \frac{-3\pm \sqrt{29}}{10}}

So, let us take x 1 = − 3 + 29 10 x_{1}=\large{\frac{-3+\sqrt{29}}{10}} and x 2 = − 3 − 29 10 x_{2}=\large{\frac{-3-\sqrt{29}}{10}}

Then, 1 x 1 + 1 x 2 = 10 − 3 + 29 + 10 − 3 − 29 = 10 ( − 3 − 29 − 3 + 29 ) 9 − 29 = 10 × ( − 6 ) ( − 20 ) = 3 \large{\frac{1}{x_1}+\frac{1}{x_2}=\frac{10}{-3+\sqrt{29}}+\frac{10}{-3-\sqrt{29}} = \frac{10(-3-\sqrt{29}-3+\sqrt{29})}{9-29}=\frac{10\times (-6)}{(-20)}=\boxed{3}}

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At the last, I didn't show the rationalization properly, so I am showing it here in the comment. First you should know the identity ( a + b ) ( a − b ) = a 2 − b 2 (a+b)(a-b)=a^2-b^2 . Here is the rationalization :

10 − 3 + 29 + 10 − 3 − 29 \large{\frac{10}{-3+\sqrt{29}}+\frac{10}{-3-\sqrt{29}}}

= 10 ( ( − 3 − 29 ) + ( − 3 + 29 ) ) ( − 3 + 29 ) ( − 3 − 29 ) =\large{\frac{10((-3-\sqrt{29})+(-3+\sqrt{29}))}{(-3+\sqrt{29})(-3-\sqrt{29})}}

= 10 ( − 3 − 29 − 3 + 29 ) ( − 3 + 29 ) ( − 3 − 29 ) =\large{\frac{10(-3-\sqrt{29}-3+\sqrt{29})}{(-3+\sqrt{29})(-3-\sqrt{29})}}

= 10 × ( − 6 ) ( − 3 ) 2 − ( 29 ) 2 =\large{\frac{10\times (-6)}{(-3)^2-(\sqrt{29})^2}}

= 10 × ( − 6 ) 9 − 29 =\large{\frac{10\times (-6)}{9-29}}

And the rest is done in the given solution.

Prasun Biswas - 7 years, 4 months ago

You could simply use Vieta's Formula.

Harsh Shrivastava - 6 years, 9 months ago

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