The discussion revolves around solving the quadratic equation \(2x^2 + 5x - k\) under the condition that its roots differ by 2. Participants explore various methods to find the value of \(k\), including the use of the quadratic formula and alternative approaches involving root relationships.
Participants generally agree on the methods to approach the problem, but there is no consensus on a single solution or method being superior. Multiple approaches are discussed without resolving which is the most effective.
Participants note that the roots are not provided, and the only information given is their difference. The discussion reflects various assumptions about the nature of the roots and the implications for solving for \(k\).
This discussion may be useful for students or individuals interested in quadratic equations, particularly those exploring relationships between roots and coefficients in algebraic expressions.
MarkFL said:Begin by stating the roots of the quadratic, using the quadratic formula...what do you have?
RTCNTC said:Discriminant?
MarkFL said: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?
RTCNTC said:The roots are not given. The roots differ by 2. This is the only data given.
MarkFL said: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
topsquark said: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
MarkFL said: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