Question

If one root of the quadratic equation a^2+bx+c=0 is 3-4i then a+b+c=

Answer 1

The correct answer is 2c which is an option (d).

Step-by-step explanation:

• Given that a,b,cϵR, a quadratic equation with real coefficients and a complex number as one of its roots would have the other root as its complex conjugate.

• As a result, one root in the given example is 3-4i.

• Therefore other roots will be 3+4i.

• Hence

−b/a =3+4i+3−4i

Hence b=−6a...(i)

And

c/a =(3+4i)(3−4i)

c=25a ...(ii)

Therefore the equation transforms to

ax2−6ax+25a=0

Or

x2−6x+25=0

Comparing with

ax2+bx+c=0 gives us

a=1 b=−6,c=25

Hence

31a+b+c

=31(1)−6+25

=31+25−6

=50

=2(25)

=2c

The complete question is ''If one root of the quadratic equation a^2+bx+c=0 is 3-4i then a+b+c= , (a)0, (b) 2a, (c) 2b, (d) 2c.

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Answer 2

Answer:

a+b+c=20

STEP -STEP SOLUTION:

If one of roots of a quadratic equation is a number with real coefficient and accomplish number then the other root is its complex conjugate.

i.e, 3+4i in the above case.

hence,

• sum of 2 roots is:

-b/a=(3+4i)+(3-4i)

• product of 2 roots is:

c/a =(3+4i)x(3-4i)

upon solving the above equation we get:

b=-6a-------(i)

c=25a------(ii)

upon putting values of (i) and (ii) in the quadratic equation.

we get;

a²+(-6a)x+25a=0

which gives us

• a=1
• b=-6
• c=25

therefore ,

a+b+c= 1+25-6

=20

(#SPJ3)