Nature of Roots

**NeF** · July 8th 2006, 02:08 PM

Hey sup got stuck with nature of roots, so sad -_-

-> Show that the equation mx^2 + 5x = 5m has real and unequal roots for all real values of m

and

-> Find the values of k for which the roots of the equation
kx^2 + 2(k-3)x + k - 5 = 0 are real. Give 3 values for k for which the roots are rational.

Thanx

**Jameson** · July 8th 2006, 02:26 PM

1) $mx^2+5x-5m=0$

Use the quadratic formula and both claims will be demonstrated.

2) Look at the equation like this: $kx^2+(2k-6)x+(k-5)$

The discriminant or, $\sqrt{b^2-4ac}$ part, determines the real/imaginary set of the roots. So in order for the whole root to be real (assuming coefficients are real as well), $\sqrt{(2k-6)^2-4(k)(k-5)}$ must be real, meaning that $(2k-6)^2-4(k)(k-5)>0$

**galactus** · July 8th 2006, 02:35 PM

Originally Posted by NeF

Hey sup got stuck with nature of roots, so sad -_-

-> Show that the equation mx^2 + 5x = 5m has real and unequal roots for all real values of m

Try using the discriminant.

$b^{2}-4ac$

$25-4(m)(-5m)$

The roots are real and unequal if the discriminant is positive.

$25+20m^{2}$ is positive for all real values of m.

**NeF** · July 8th 2006, 02:41 PM

[QUOTE=galactus][QUOTE=NeF]Hey sup got stuck with nature of roots, so sad -_-

-> Show that the equation mx^2 + 5x = 5m has real and unequal roots for all real values of m

Try using the discriminant.

$b^{2}-4ac$

$25-4(m)(-5m)$

The roots are real and unequal if the discriminant is positive.

$25+20m^{2}$ is positive for all real values of m.

I got as far as $25+20m^{2}$ would the next step be $5+4m^{2}$ ?

**galactus** · July 8th 2006, 03:06 PM

You can do that if you like, but it's done. The square assures you that all real values of m are positive.

**Quick** · July 8th 2006, 05:03 PM

Originally Posted by NeF

Originally Posted by galactus

Originally Posted by NeF

Hey sup got stuck with nature of roots, so sad -_-

-> Show that the equation mx^2 + 5x = 5m has real and unequal roots for all real values of m

I got as far as $25+20m^{2}$ would the next step be $5+4m^{2}$ ?

Actually, no, becuase $25+20m^2\neq5+4m^2$ however, it doesn't actually matter becuase you've already shown that the discriminant is positive.

**topsquark** · July 8th 2006, 05:49 PM

Originally Posted by Jameson

1) $mx^2+5x-5m=0$

Use the quadratic formula and both claims will be demonstrated.

Though technically we have to exclude m = 0 from this statement...

-Dan

**NeF** · July 9th 2006, 11:25 AM

Thanks for your help guys. One thing I realised is that I should post how far I got with each sum when posting questions.

This is as far as I got with the second* question. Please let me know if I am doing something wrong cause all I did was random stuff x/

$kx^{2} + 2(k-3)x + k - 5 = 0 \sqrt{b^2-4ac} (2k-6)^2-4(k)(k-5)>0 (2k-6)(2k-6) - 4(k^2-5k) > 0 4k^2-12k-12k+36-4k^2+20k>0 -4k = -36 divide by -4 k = 9$

$ kx^2 + 2(k-3)x + k - 5 = 0$
$9x^2+2(9-3)x+9-5 = 0$

$9x^2+12x+4=0 (3x+2)(3x+2)=0$
$(3x+2)^2 = 0$

$Then take (3x+2) = 0$
$x = -2/3$
and Sub the x and k values into
$ kx^2 + 2(k-3)x + k - 5 = 0$
$(9)(-2/3)^2+2(9-3)-2/3+9-5=0$

So all 3 rational values are $9 ; (3x+2)^2 ; 0$
Is that correct and btw the sum is worth 12 marks.
Sorry if I come across as a complete doos

**malaygoel** · July 10th 2006, 03:35 AM

Originally Posted by NeF

Thanks for your help guys. One thing I realised is that I should post how far I got with each sum when posting questions.

This is as far as I got with the second* question. Please let me know if I am doing something wrong cause all I did was random stuff x/

$kx^{2} + 2(k-3)x + k - 5 = 0 \sqrt{b^2-4ac} (2k-6)^2-4(k)(k-5)>0 (2k-6)(2k-6) - 4(k^2-5k) > 0 4k^2-12k-12k+36-4k^2+20k>0 -4k = -36 divide by -4 k = 9$

$ kx^2 + 2(k-3)x + k - 5 = 0$
$9x^2+2(9-3)x+9-5 = 0$

$9x^2+12x+4=0 (3x+2)(3x+2)=0$
$(3x+2)^2 = 0$

$Then take (3x+2) = 0$
$x = -2/3$
and Sub the x and k values into
$ kx^2 + 2(k-3)x + k - 5 = 0$
$(9)(-2/3)^2+2(9-3)-2/3+9-5=0$

So all 3 rational values are $9 ; (3x+2)^2 ; 0$
Is that correct and btw the sum is worth 12 marks.
Sorry if I come across as a complete doos

Discriminant is $36 - 4k$
Roots are real for $k\leq9$
For rational roots $36 - 4k$ is a perfect square.

Keep Smiling
Malay

**Quick** · July 10th 2006, 04:21 AM

This is an incorrect statement...

Originally Posted by NeF

$ 4k^2-12k-12k+36-4k^2+20k>0 -4k = -36 divide by -4 k = 9$

There are two things wrong with your work, you can't change > into = whenever you feel like it, and it should say >= 0

here is my work...

$4k^2-12k-12k+36-4k^2+20k\geq0$

$-4k+36\geq0$

$-4k\geq-36$

$k\leq9$ note:when multiplying or dividing by a negative number, you need to flip the signs.

**NeF** · July 11th 2006, 11:31 AM

Awesome thanks for pointing out my mistakes

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