Question

If a and Bare zeroes of the quadratic polynomial
4x² + 4x + 1, then form a quadratic polynomial whose zeroes are 2a and 2B.

Answer 1

Answer:

Firstly use sridharacharya's rule to solve this first quadratic equation .

Now, according to that , alpha=(-4+((4²-4×4×1))^1/2)/2×4= -4/8=-1/2

so, Beta =-4-0/8=-1/2

now, here D=b²-4ac

so, let's arrange this for second quadratic polynomial........

ax²+bx+c

x²+(alpha+beta)x-alpha ×beta

x²+(2×-1/2 + 2×-1/2)x-(-1/2×-1/2)

x²+ -2x-1/4

multiply this quadratic polynomial with 4

so, 4x²-8x-1

hope it helps!!!!

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