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To find the nature of roots of `36x^2+12x-1=0`
use the formula for the discriminant (delta):
`Delta = b^2-4ac` where a=36, b=12 and c= -1
`therefore Delta= 12^2-4(36)(-1)`
`Delta = 288`
Now to analyse the discriminant:
We know that 288 is greater than zero (>0) and it is not a perfect square.
Therefore the roots will be real, irrational and unequal.
To now solve it using the quadratic formula, note that we have already solved it partially so do not need to redo the contents of the square root which we know to be 288 . The formula is:
`therefore x=(-12 +-sqrt(288))/(2a)`
`therefore x=0.069` or `x= -0.402` (NB: These numbers have been rounded off as we know that they are actually irrational)
Ans: Roots are real, unequal and irrational to the values of x=0.069 and x=- 0.402
`36x^2 + 12x -1 = 0 ` use the formula `b^2-4ac` to find the discriminant
`a=36` `b=12 ` `c=-1`
`144+144=288` so the discriminant is 288 since it is bigger than 0 it means this problem has 2 real solutions to find the
roots use the quadratic formula `(-b+-sqrt(b^2-4ac))/(2a)`
set the problems up differently
`a!=0 and a,b,c in R`
(i) `Delta>0` then roots of the given quadratic equation are real and unequal ( may be rational or irrational).
(ii)`Delta=0` , then roots of the given quadratic equation are real and equal ( may be rational or irrational).
(iii) `Delta<0` then roots of the given quadratic equation are not real .
Given equation is
a=36,b=12 and c=-1
So roots are real and unequal.
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