# Thread: Quadratic Equation: Condition for Roots Given, Coefficient Must be Figured Out

1. ## Quadratic Equation: Condition for Roots Given, Coefficient Must be Figured Out

The question is:

If one of the roots of x<sup>2</sup> + ax + 4 = 0 is twice the other root, then the value of 'a' is?

(A) -3(2)<sup>0.5</sup>
(B) 8(2)<sup>0.5</sup>
(C) (2)<sup>0.5</sup>
(D) -2(2)<sup>0.5</sup>

2. If a quadratic equation $x^2+ax+b=0$ has roots $x_1,x_2$, then $x^2+ax+b=(x-x_1)(x-x_2)$. Also, two polynomials are equal iff their respective coefficients are equal.

According to my calculations, the answer is "none of the above" because one variant is only a partial, not complete, answer.

In typing polynomials in ASCII, it is customary to indicate power using ^, e.g.: x^2 + ax + 4. You can also use LaTeX: type $$x^2+ax+4$$ and $$\sqrt{2}$$ to get $x^2+ax+4$ and $\sqrt{2}$.

The question is:

If one of the roots of x<sup>2</sup> + ax + 4 = 0 is twice the other root, then the value of 'a' is?

(A) -3(2)<sup>0.5</sup>
(B) 8(2)<sup>0.5</sup>
(C) (2)<sup>0.5</sup>
(D) -2(2)<sup>0.5</sup>

1. Solve the equation

$x^2+ax+4=0$ for x.

You should come out with:

$x_1 = -\frac a2-\sqrt{\frac{a^2}4-4}~\vee~x_1 = -\frac a2+\sqrt{\frac{a^2}4-4}$

2. According to the text of the peoblem you know that

$2x_1 = x_2$ . That means:

$2\cdot \left(-\frac a2-\sqrt{\frac{a^2}4-4}\right)= -\frac a2+\sqrt{\frac{a^2}4-4}$

Solve for a.

3. Btw I've got 2 different values for a. But please check my calculations!

4. Yet another variation: Let $x_1$ be one of the roots- the other is $2x_1$. Then we have $(x- x_1)(x- 2x_1)= x^2- (x_1+ 2x_1)+ 2x_1^2= x^2- 3x_1x+ 2x_1^2= x^2+ ax+ 4$
Now you know that $2x_1^2= 4$ and $a= -3x_1$. Yes, there are two possible values for a but only one of them is one of the choices.

If you can't write $x^2$, then please at least use x^2 so that what you post is more easily read etc.