If the D of 3x²+2x+a=0 is double the D of x²-4x+2,then find the value of a??
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D of 1st eqn will be
=b^2-4ac=2^2-4*3*a=4-12a
D of 2nd eqn will be
=(-4)^2-4*1*2=16-8=8
it is given that D of 1st eqn is double than D of 2nd eqn.
so,
4-12a=8
=>12a=4-8=>12a=-8
=>a=-8/12
=>a=-2/3
=b^2-4ac=2^2-4*3*a=4-12a
D of 2nd eqn will be
=(-4)^2-4*1*2=16-8=8
it is given that D of 1st eqn is double than D of 2nd eqn.
so,
4-12a=8
=>12a=4-8=>12a=-8
=>a=-8/12
=>a=-2/3
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