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Find signs (+/-) of the roots of a quadratic equation

Expert replies
by clawhammer » Thu Nov 04, 2010 10:20 pm
There is a method to determine if:

a) both roots are positive
b) booth roots are negative
c) one root is positive and another is negative

- from the signs in the quadratic equation, without solving the equation.

For example:

x^2 -5x - 150 = 0 - I need to know if there is only one positive root for this equation. How? :(
So as long as I know there's just one positive value, i dont need to solve to determine this is sufficient to know x. (where x can only be positive)

Can anyone tell me?
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Source: — Data Sufficiency |

by fskilnik@GMATH » Fri Nov 05, 2010 4:39 am
clawhammer wrote:There is a method to determine if:

a) both roots are positive
b) booth roots are negative
c) one root is positive and another is negative

- from the signs in the quadratic equation, without solving the equation.

For example:

x^2 -5x - 150 = 0 - I need to know if there is only one positive root for this equation. How? :(
So as long as I know there's just one positive value, i dont need to solve to determine this is sufficient to know x. (where x can only be positive)

Can anyone tell me?
Sure!

If the 2o degree equation has (real) roots, that is, if the discriminant ("b^2-4ac") is non-negative, the sum of the roots is always -b/a and their product is always c/a therefore:

If product is positive, both have the same sign and, if so, if their sum is positive, they are both positive (and they may be equal...)

If the product is negative, opposite signs, the sum dictates the greater absolute value...

Hope it helps.

Regards,
Fabio.

In your example: x^2 -5x - 150 = 0

b^2-4ac = (-5)^2 -4(1)(-150) is positive, good: two (real and) distinct roots!!

Sum: 5
Product : -150,

Therefore opposite signs, the positive root with greater absolute value...
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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reply

by harsh8686 » Mon Dec 10, 2018 4:13 pm
be vary of extraneous roots. Do go through this link.
https://magoosh.com/gmat/2013/gmat-math ... -radicals/
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