The question is: determine the value(s) of m in each equation so that the two roots are equal. My teacher told me to use a discriminant. What's that?

$\displaystyle x^2-6x+m=o$
$\displaystyle 4x^2+mx+25$

Determine the value(s) of n in each equation such that one root is triple the other root.

$\displaystyle 3x^2-4x+n=0$
$\displaystyle 4x^2+nx+27=0$

Help would be GREATLY appreaciated!!!

Thank you!

2. Originally Posted by nathan02079
The question is: determine the value(s) of m in each equation so that the two roots are equal. My teacher told me to use a discriminant. What's that?

$\displaystyle x^2-6x+m=o$
$\displaystyle 4x^2+mx+25$

Determine the value(s) of n in each equation such that one root is triple the other root.

$\displaystyle 3x^2-4x+n=0$
$\displaystyle 4x^2+nx+27=0$

Help would be GREATLY appreaciated!!!

Thank you!

$\displaystyle ax^2+bx+c=0$

has the solutions:

$\displaystyle x = -\dfrac b{2a} \pm \dfrac1{2a} \sqrt{\underbrace{b^2-4ac}_{discriminant}}$

The discriminant is the term under the root-sign.

According to your first question you get:

$\displaystyle x^2-6x+m=0~\implies~x=3\pm\sqrt{9-m}$

To get only one result the discriminant must be zero. Therefore m must be 9. Then both roots are 3.

Do the second example in just the same way.

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Solve $\displaystyle 3x^2-4x+n=0$ for x:

$\displaystyle x=-\dfrac{-4}{2\cdot 3}\pm \dfrac1{6} \sqrt{16-4 \cdot 3 \cdot n}$

$\displaystyle x_1=\dfrac23-\dfrac16\sqrt{16-12n}~\vee~x_2=\dfrac23+\dfrac16\sqrt{16-12n}$

According to the question you know that $\displaystyle x_2 = 3\cdot x_1$:

$\displaystyle \dfrac23+\dfrac16\sqrt{16-12n} = 3\cdot \left( \dfrac23-\dfrac16\sqrt{16-12n} \right)$

$\displaystyle \dfrac23+\dfrac16\sqrt{16-12n} = 2-\dfrac12\sqrt{16-12n}$

$\displaystyle \dfrac23\sqrt{16-12n} = \dfrac43$

$\displaystyle \sqrt{16-12n} = 2$

$\displaystyle 16-12n = 4~\implies~\boxed{n=1}$
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