The equation can have two forms $$ax^2+2bx-c+0$$ Or $$ax^2-2bx-c=0$$ both discriminants of the equation are $${4b^2+4ac}$$ which is a positive value so roots are real. Then there should be four roots of the given equation, but answer says there are only two. How is that possible?
Asked
Active
Viewed 590 times
1
-
Each form of the equation holds under a condition for the root. The first form under the condition that $x\geq0$, while the second form holds under the condition that $x<0$. The roots found for each equation need to be filtered according to whether they satisfy or not the corresponding condition. – conditionalMethod Nov 17 '19 at 05:05
-
1For example, the first form has the solution $\frac{-2b-\sqrt{4b^2+4ac}}{2a}<0$, which is negative. But the first form is equivalent to the original equation only for $x\geq0$. Therefore, this one is not a solution of the original equation. – conditionalMethod Nov 17 '19 at 05:06
-
@conditionalMethod how do you know that the first from is $<0$. – Aditya Nov 17 '19 at 05:32
2 Answers
2
Since $x^2 = |x|^2$ the given equation is same as:$$a|x|^2+2b|x|-c=0$$
This is a quadratic in $|x|$ and the quadratic formula gives $$|x|=(-b\pm \sqrt{b^2+ac})/a$$
However you must discard the negative root $(-b-\sqrt{b^2+ac})/a$ because $|x|$ is nonnegative.
$$|x| = (-b+\sqrt{b^2+ac})/a\Rightarrow x = \pm(-b+\sqrt{b^2+ac})/a$$
AgentS
- 12,195
1
I got $$x_{1,2}=-\frac{b}{a}\pm\sqrt{\frac{b^2}{a^2}+\frac{c}{a}}$$ or $$x_{1,2}=\frac{b}{a}\pm\sqrt{\frac{b^2}{a^2}+\frac{b}{a}}$$ only $x_1$ (in both cases) are solutions.
Dr. Sonnhard Graubner
- 95,283
-
-
-
-
$$x_2=-\frac{b}{a}-\sqrt{\frac{b^2}{a^2}+\frac{c}{a}}<0$$ and $$x_2=\frac{b}{a}-\sqrt{\frac{b^2}{a^2}+\frac{c}{a}}<0$$ since $a,b,c>0$ – Dr. Sonnhard Graubner Nov 17 '19 at 05:38
-
Yes, but my question was why can’t $x_2$ be negative, since it’s getting squared or $| |$ anyway – Aditya Nov 17 '19 at 07:31