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Bunuel
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Kris88
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Kris88
I understand the point of the article, but what would be the solution for " x^2 – 5x – 5 = 0" where it will be positive and negative?

tdotmba
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I understand the point of the article, but what would be the solution for " x^2 – 5x – 5 = 0" where it will be positive and negative?

You would have to use the quadratic equation if you actually wanted to solve it.
Your two solutions of x would equal [(5+ √45) / 2] & [(5 - √45) / 2]

approx: (x - 5.854) (x + 0.854 )

You actually don't have to solve. Sum of roots = 5 and product is -ve. So definitely one of them is +ve and other one is -ve and +ve root is more than -ve root by 5.
And as quadratic equations are only defined for real numbers, we will be having only real number solution.
So statement 1 will not be sufficient.

So statement 2 will be required to solve the DS question. Answer will be C)
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Kris88
I understand the point of the article, but what would be the solution for " x^2 – 5x – 5 = 0" where it will be positive and negative?

tdotmba
Kris88
I understand the point of the article, but what would be the solution for " x^2 – 5x – 5 = 0" where it will be positive and negative?

You would have to use the quadratic equation if you actually wanted to solve it.
Your two solutions of x would equal [(5+ √45) / 2] & [(5 - √45) / 2]

approx: (x - 5.854) (x + 0.854 )

You actually don't have to solve. Sum of roots = 5 and product is -ve. So definitely one of them is +ve and other one is -ve and +ve root is more than -ve root by 5.
And as quadratic equations are only defined for real numbers, we will be having only real number solution.
So statement 1 will not be sufficient.

So statement 2 will be required to solve the DS question. Answer will be C)


Kris88 said he understood the article but just wanted to know what the two solutions would be anyways (I suppose in case a quadratic question appeared in PS to solve for actual value rather than DS) :)
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