Calculus or Algebra?

Algebra Level 4

If one root of the equation ( I m ) x 2 + I x + 1 = 0 (I-m)x^2+Ix+1=0 is double of the other and is real, then the greatest value of m is equal to???


The answer is 1.125.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Brian Riccardi
Jun 25, 2015

If we solve the the equation in x x we find two roots:

x 1 = I Δ 2 ( I m ) , x 2 = I + Δ 2 ( I m ) , Δ = I 2 4 ( I m ) x_1=\frac{-I-\sqrt{\Delta}}{2(I-m)}, x_2=\frac{-I+\sqrt{\Delta}}{2(I-m)}, \Delta=I^{2}-4(I-m)

Without loss of generalities:

2 x 1 = x 2 2x_1=x_2

2 ( I Δ ) = I + Δ 2( -I-\sqrt{\Delta} )=-I+\sqrt{\Delta}

m = 9 I 2 I 2 9 m=\frac{9I-2I^{2}}{9}

Now we can study m ( I ) m(I) as a function of I I and find a m a x max in I = 9 4 I=\frac{9}{4}

The solution of the problem is m ( 9 4 ) = 9 8 m(\frac{9}{4})=\frac{9}{8}

1.125 \boxed{1.125}

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...