f(x)=4√3xsquare+5x-2√3 verify zeros and cofficients
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Answer:
f(x)=4√3x^2+5x-2√3
*product=4√3*-2√3=-24
*sum=5x
SPLITTING THE MIDDLE TERM:
4√3x^2+8x-3x-2√3
4x(√3x-2)+√3(√3x-2)
(4x+√3)(√3x-2)
therefore the zeroes are,
-√3/4 and 2/√3
let alpha=-√3/4 and beta=2/√3,
we know,
alpha + beta=-b/a and alpha*beta=c/a
ie,-√3/4+2/√3=-5/4√3 -√3/4*2/√3=-2√3/4√3
-3+8/4√3=-5/4√3 -2/4=-1/2
-5/4√3=-5/4√3 -1/2=-1/2
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