for $x \in \mathbb{R}$ find the number of solutions in $3x^2 + 4|x^2 - 1| + x - 1 = 0$. Why we are considering negative values here?

Question

I was solving some practice problems, from a booklet of math, and one of the problems is like this

for $x \in \mathbb{R}$ find the number of roots in the eq . $$3x^2 + 4|x^2 - 1| + x - 1 = 0$$

I know I have find the value of $\Delta = b^2 - 4ac$, and based on that I can say something about $\#$ of real roots.

But I looked at 3 help books, and some online resources as well, and the way they have solved this is they consider two cases - first where $x^2 - 1 \geq 0 $ and the other is where $x^2 - 1 < 0$

I don't understand why these books are considering the negative part, whatever the value is inside it will be positive after taking absolute value so I think the solution should be equal to the number of real roots for the equation $$3x^2 +4(x^2-1) +x -1 = 0$$ $$3x^2 + 4x^2 - 4 + x - 1= 0$$ $$7x^2 + x - 5 = 0$$

$$\Delta = 1^2 - 4(7)(-5) = 141$$

so give eq. has 2 real roots.

I can't provide you the textbook, but in an online website the problem has been solved in same way which can be found here - https://byjus.com/question-answer/for-x-in-mathbb-r-the-number-of-real-roots-of-the-equation-3x-2-3/

Answer 1

There are two possible cases: $\,|x|<1\,$ or $\,|x|\geqslant1\,.$

If $\,|x|<1\,,\,$ then

$3x^2+4\left|x^2-1\right|+x-1=3x^2-4\left(x^2-1\right)+x-1=\\=-x^2+x+3\geqslant-x^2-|x|+3=-|x|\big(|x|+1\big)+3>\\>-2+3=1\,.$

If $\,|x|\geqslant1\,,\,$ then

$3x^2+4\left|x^2-1\right|+x-1=3x^2+4\left(x^2-1\right)+x-1=\\=7x^2+x-5=\dfrac12\big(x+1\big)^2+\dfrac{13}2x^2-\dfrac{11}2\geqslant\\\geqslant\dfrac{13}2x^2-\dfrac{11}2\geqslant\dfrac{13}2-\dfrac{11}2=1\,.$

Consequently, it results that

$3x^2+4\left|x^2-1\right|+x-1\geqslant1\quad$ for any $\;x\in\Bbb R\;,$

hence the equation

$3x^2+4\left|x^2-1\right|+x-1=0$

does not have any real solution.

Answer 2

It follows from the definition of the absolute value function. $$ |x| = \begin{cases} x & x \ge 0 \\ -x & x < 0 \end{cases}$$

Thus $$ |x^2-1| = \begin{cases} x^2-1 & x^2-1 \ge 0 \\ -(x^2-1) & x^2-1 < 0 \end{cases}$$

The absolute value of any real number is indeed always non-negative, but that doesn't mean the argument to the absolute value function is always non-negative.

Take for example, $x = 0.5$ $$ \begin{align} |x^2-1| &= |0.25 - 1| = 0.75 \\ x^2 - 1 &= 0.25 - 1 = -0.75 \\ -(x^2 - 1) &= -(0.25 - 1) = 0.75 \end{align} $$

As you can see, $|x^2-1| = -(x^2 - 1)$ for such a value of $x$

Your mistake is the assumption that $|x^2 - 1| = x^2 - 1$ is true for any $x$. When $x^2 - 1$ is negative, you can clearly see that this cannot be true as the absolute value of a real number cannot be negative.