Given ax2+bx+c=0 ⇒ax2=−bx−c Now, consider y=4ax2+3bx+2c=4[−bx−c]+3bx+2c=−4bx−4c+3bx+2c=−bx−2c Since, this curve intersects x-axis ∴ put y=0, we get −bx−2c=0⇒−bx=2c⇒x=b−2c Thus, given curve intersects x-axis at exactly one point.
JEE Main 2012 — Mathematics Algebra
If a,b,c∈R and 1 is a root of equation ax2+bx +c=0, then the curve y=4ax2+3bx+2c,a=0 intersect x-axis at
Held on 26 May 2012 · Verified 6 Jul 2026.
two distinct points whose coordinates are always rational numbers
no point
exactly two distinct points
exactly one point
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