There is more than one root of all evil

Algebra Level 2

Given that $x_{1}$ and $x_{2}$ are the roots to the following polynomial $4x^{2}-32x-k=0,$ and $x_{1}$ + $x_{2}$ =8, and | $x_{1}$ $x_{2}$ | = 345, find the value of $|k|$ .

The answer is 1380.

3 solutions

Joshua Ong
May 18, 2014

Based on Vieta's theorem, the product of the roots $x_1$ and $x_2$ is equivalent to $\frac{c}{a}$ , where c and a are the terms in the equation $ax^2+bx+c$ . In the equation in the problem, $c$ and $a$ are $k$ and $4$ respectively. Since $\frac{k}{4}$ is $345$ , thus $k=345 \times 4=\boxed{1380}$

this is false, we have |k|=1380, not k=1380. and c is actually -4. and it is actually $|\frac{k}{4}|=345$ , not $\frac{k}{4}=345$ .

mathh mathh - 6 years, 11 months ago

William Isoroku
Aug 1, 2014

Product of the roots is c/a

Syed Shahabudeen
May 15, 2014

In quadratic equation the relation between the roots states that : x1+x2=32/4 and x1 x2=k/4 this implies that k/4=345, therefore k=345 4=1380

same mistake as the other solution - don't forget we have the absolute value signs, which change a lot of things.

mathh mathh - 6 years, 11 months ago

There is more than one root of all evil

The answer is 1380.

3 solutions

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