Rajkiranmareedu wrote:
Bunuel wrote:
subhashghosh wrote:
Hi
I have a doubt in this question :
Given that x^4 - 25x^2 = -144 , which of the following is NOT a sum of two possible values of x ?
Official Answer - D
The roots are x = 4 or -4, and -3 or +3
These values are considered for the sum,
-7 = -4+ -3
-1 = -4 + 3
0 = -4 + 4 or -3 + 3
7 = 4 + 3
But why was 4 -3 not considered ?
Regards,
Subhash
Subhash, please read and follow: how-to-improve-the-forum-search-function-for-others-99451.html So please:
Provide answer choices for PS questions.Original question is:
Given that x^4 – 25x^2 = -144, which of the following is NOT a sum of two possible values of x? A. -7
B. -1
C. 0
D. 3
E. 7
Factor
\(x^4-25x^2+144=0\) --> \((x^2 - 16)*(x^2 - 9)=0\) --> \(x^2=16\) or \(x^2=9\) (alternately you could solve \(x^4-25x^2+144=0\) for \(x^2\) to get the same values for it) --> \(x=4\) or \(x=-4\) or \(x=3\) or \(x=-3\).
All but option D. could be expressed as the sum of two roots: A. 7=-4-3; B. -1=-3+4; C. 0=3-3 (or 0=4-4); E. 7=3+4.
Answer: D.
As for your question: there are no enough answer choices to show all possible values of a sum of two roots, so there are 4 possible values for a sum and one which is not.
Hope it's clear.
How did u calculate the highlighted part.
To calculate the roots of a quadratic equation ax^2+bx+c=0
you can use the formula
roots =(-b
+sqrt(b^2-4ac))/2a...and .....(-b
-sqrt(b^2-4ac))/2a
now in this x^4-25x^2+144=0===>just for convenience put x^2= X
therefore equation becomes: X^2-25X+144=0
NOW using the formula of roots
X=(25+sqrt(25^2-4*1*144))/2 and X=(25-sqrt(25^2-4*1*144))/2
On simplifying
X=16...AND X=9
now replacing back X WITH x
we get x^2=16 and x^2=9
therefore four roots are x=+4,-4,+3,-3
hope it helps...
SKM
hope it h