y = 4x^2 + 4x 8
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14 Feb 2024, 23:03
to solve \(y=4x^2+4x-8\) look for two numbers that add up to \(+4\) which is the coefficient of the linear term and multiply to \(-32\) which is the product of the constant term and the coefficient of the quadratic term
The two numbers clearly are \(+8\) and \(-4\)
So we rewrite \(4x^2+4x-8=0\) as
\(4x^2+8x-4x-8=0 \)
\(2x(2x+4) -2(2x+4) =0\)
\((2x+4)(2x-2)=0\)
\(2x+4 = 0\)
\(2x=-4\)
\(x=-4/2\)
\(x=-2\)
\(2x-2 =0\)
\(2x = 2\)
\(x=2/2\)
\(x=1\)
The roots of the equation are \(-2\) and \(1\).
Quantity A = the Smaller root \(= -2\)
Quantity B \(= -2\)
BOTH quantities are EQUAL
The answer is C