MHB How can I solve the equation to find the value of k when the roots differ by 2?

[mathdad]

The roots of the equation 2x^2+ 5x – k differ by 2. Find the value of k.

Can someone get me started? I found this question online and find it interesting.

 

[MarkFL]

Begin by stating the roots of the quadratic, using the quadratic formula...what do you have?

 

[mathdad]

Begin by stating the roots of the quadratic, using the quadratic formula...what do you have?

Discriminant?

 

[MarkFL]

Discriminant?

No, the roots themselves. Recall, if:

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

then:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

What are the roots of the given quadratic?

 

[mathdad]

No, the roots themselves. Recall, if:

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

then:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

What are the roots of the given quadratic?

The roots are not given. The roots differ by 2. This is the only data given.

 

[MarkFL]

The roots are not given. The roots differ by 2. This is the only data given.

True, the roots are no given, but you can compute them...we should begin by equating the given quadratic to zero:

$$2x^2+5x–k=0$$

Using the quadratic formula, we find the roots to be:

$$x=\frac{-5\pm\sqrt{5^2-4(2)(-k)}}{2(2)}=\frac{-5\pm\sqrt{25+8k}}{4}$$

Now, if the roots are to differ by 2, then take the larger root, subtract the smaller, and set that difference equal to 2:

$$\frac{-5+\sqrt{25+8k}}{4}-\frac{-5-\sqrt{25+8k}}{4}=2$$

Now solve this for $k$...

Another approach would be to write:

$$2x^2+5x–k=2(x-(r+2))(x-r)=2x^2-4(r+1)x+2r(r+ 2)$$

Equating coefficients, we obtain:

$$-4(r+1)=5$$

$$k=-2r(r+ 2)$$

Solve the first to get $r$, and then use that value of $r$ in the second to determine $k$. :D

 

[mathdad]

True, the roots are no given, but you can compute them...we should begin by equating the given quadratic to zero:

$$2x^2+5x–k=0$$

Using the quadratic formula, we find the roots to be:

$$x=\frac{-5\pm\sqrt{5^2-4(2)(-k)}}{2(2)}=\frac{-5\pm\sqrt{25+8k}}{4}$$

Now, if the roots are to differ by 2, then take the larger root, subtract the smaller, and set that difference equal to 2:

$$\frac{-5+\sqrt{25+8k}}{4}-\frac{-5-\sqrt{25+8k}}{4}=2$$

Now solve this for $k$...

Another approach would be to write:

$$2x^2+5x–k=2(x-(r+2))(x-r)=2x^2-4(r+1)x+2r(r+ 2)$$

Equating coefficients, we obtain:

$$-4(r+1)=5$$

$$k=-2r(r+ 2)$$

Solve the first to get $r$, and then use that value of $r$ in the second to determine $k$. :D

This problem is more involved than the other question in terms of solving for n. Your help and guidance is greatly appreciated.

 

[topsquark]

As MarkFL commented, you need to solve the following for k:
$$\frac{-5+\sqrt{25+8k}}{4}-\frac{-5-\sqrt{25+8k}}{4}=2$$

[math](-5+\sqrt{25+8k}) - (-5-\sqrt{25+8k}) = 4 \cdot 2 = 8[/math]

[math](-5 - (-5)) + (\sqrt{25 + 8k} - - \sqrt{25 + 8k}) = 8[/math]

[math]2~\sqrt{25 + 8k} = 8[/math]

Can you finish from here?

-Dan

 

[mathdad]

As MarkFL commented, you need to solve the following for k:
$$\frac{-5+\sqrt{25+8k}}{4}-\frac{-5-\sqrt{25+8k}}{4}=2$$

[math](-5+\sqrt{25+8k}) - (-5-\sqrt{25+8k}) = 4 \cdot 2 = 8[/math]

[math](-5 - (-5)) + (\sqrt{25 + 8k} - - \sqrt{25 + 8k}) = 8[/math]

[math]2~\sqrt{25 + 8k} = 8[/math]

Can you finish from here?

-Dan

I got it from here.

- - - Updated - - -

True, the roots are no given, but you can compute them...we should begin by equating the given quadratic to zero:

$$2x^2+5x–k=0$$

Using the quadratic formula, we find the roots to be:

$$x=\frac{-5\pm\sqrt{5^2-4(2)(-k)}}{2(2)}=\frac{-5\pm\sqrt{25+8k}}{4}$$

Now, if the roots are to differ by 2, then take the larger root, subtract the smaller, and set that difference equal to 2:

$$\frac{-5+\sqrt{25+8k}}{4}-\frac{-5-\sqrt{25+8k}}{4}=2$$

Now solve this for $k$...

Another approach would be to write:

$$2x^2+5x–k=2(x-(r+2))(x-r)=2x^2-4(r+1)x+2r(r+ 2)$$

Equating coefficients, we obtain:

$$-4(r+1)=5$$

$$k=-2r(r+ 2)$$

Solve the first to get $r$, and then use that value of $r$ in the second to determine $k$. :D

Yes, I got it now. Thank you for taking an interest in my textbook questions. Keep in mind that I will only post questions in this forum after trying to solve each one on my own several times first.