Question

$\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Given:-}}}}}}}$

Solve for x:-

(4a - 15) x² + 2a |x| + 4 = 0.

@Takenname, mods, stars and other users help me.

$\green{\textsf{Class:-}}$ 11th

$\green{\textsf{Topic:-}}$ Moduless function

$\red{\textsf{Don't Spam}}$​

Answer 1

ANSWER:

Given:

• (4a - 15) x² + 2a |x| + 4 = 0

To Do:

• Solve for x

Solution:

We are given that,

$:\implies\sf(4a-15)x^2+2a|x|+4=0$

We know that,

⇒ |y|² = y²

So,

$:\implies\sf(4a-15)x^2+2a|x|+4=0$

We can rewrite it as,

$:\implies\sf(4a-15)|x|^2+2a|x|+4=0$

Let, |x| be t.

So,

$:\implies\sf(4a-15)t^2+2at+4=0$

We know that, for a quadratic equation, Ax² + Bx + C = 0, by Quadratic formula,

$:\implies\sf x=\dfrac{-B\pm\sqrt{B^2-4AC}}{2A}$

We have,

$:\implies\sf(4a-15)t^2+2at+4=0$

On applying Quadratic Formula,

$:\implies\sf t=\dfrac{-B\pm\sqrt{B^2-4AC}}{2A}$

Here,

⇒ A = (4a - 15), B = 2a and C = 4.

So,

$:\implies\sf t=\dfrac{-(2a)\pm\sqrt{(2a)^2-4(4a-15)(4)}}{2(4a-15)}$

$:\implies\sf t=\dfrac{-2a\pm\sqrt{4a^2-4(16a-60)}}{2(4a-15)}$

$:\implies\sf t=\dfrac{-2a\pm\sqrt{4a^2-64a+240}}{2(4a-15)}$

We can write it as,

$:\implies\sf t=\dfrac{-2a\pm\sqrt{4(a^2-16a+60)}}{2(4a-15)}$

$:\implies\sf t=\dfrac{-2a\pm2\sqrt{a^2-16a+60}}{2(4a-15)}$

Taking 2 common,

$:\implies\sf t=\dfrac{2\!\!\!/\,(-a\pm\sqrt{a^2-16a+60})}{2\!\!\!/\,(4a-15)}$

So,

$:\implies\sf t=\dfrac{-a\pm\sqrt{a^2-16a+60}}{4a-15}$

Now, we'll split the middle term in the quadratic equation mentioned,

$:\implies\sf t=\dfrac{-a\pm\sqrt{a^2-10a-6a+60}}{4a-15}$

$:\implies\sf t=\dfrac{-a\pm\sqrt{a(a-10)-6(a-10)}}{4a-15}$

So,

$:\implies\sf t=\dfrac{-a\pm\sqrt{(a-10)(a-6)}}{4a-15}$

We had substituted |x| for t.

So,

$:\implies\sf |x|=\dfrac{-a\pm\sqrt{(a-10)(a-6)}}{4a-15}$

Now, we know that,

⇒ |y| = +y, -y.

So,

$:\implies\sf \pm x=\dfrac{-a\pm\sqrt{(a-10)(a-6)}}{4a-15}$

1) For |x| = +x.

$:\implies\sf +x=\dfrac{-a\pm\sqrt{(a-10)(a-6)}}{4a-15}$

So,

$:\implies\bf x=-\dfrac{a+\sqrt{(a-10)(a-6)}}{4a-15}$

And,

$:\implies\bf x=-\dfrac{a-\sqrt{(a-10)(a-6)}}{4a-15}$

2) For |x| = -x.

$:\implies\sf -x=\dfrac{-a\pm\sqrt{(a-10)(a-6)}}{4a-15}$

$:\implies\sf x=-\dfrac{-a\pm\sqrt{(a-10)(a-6)}}{4a-15}$

$:\implies\sf x=\dfrac{a\pm\sqrt{(a-10)(a-6)}}{4a-15}$

So,

$:\implies\bf x=\dfrac{a+\sqrt{(a-10)(a-6)}}{4a-15}$

And,

$:\implies\bf x=\dfrac{a-\sqrt{(a-10)(a-6)}}{4a-15}$

Therefore, the possible values of x, are:

$\:\:\:\bullet\:\:\:\bf x=-\dfrac{a+\sqrt{(a-10)(a-6)}}{4a-15}$

$\:\:\:\bullet\:\:\:\bf x=-\dfrac{a-\sqrt{(a-10)(a-6)}}{4a-15}$

$\:\:\:\bullet\:\:\:\bf x=\dfrac{a+\sqrt{(a-10)(a-6)}}{4a-15}$

$\:\:\:\bullet\:\:\:\bf x=\dfrac{a-\sqrt{(a-10)(a-6)}}{4a-15}$

Answer 2

The answer made by MrImpeccable is correct, however, I found solutions do not exist for some values of $a$. Here is your answer.

Solution.

Note that $|x|^2=x^2$. We can solve for $|x|$,

$\implies |x|=\dfrac{-a\pm\sqrt{a^2-4(4a-15)} }{4a-15}$

$\implies |x|=\dfrac{-a\pm \sqrt{a^2-16a+60} }{4a-15}$

The value of $|x|$ is restricted to a positive number or 0.

Let the solutions to $(4a-15)|x|^2+2a|x|+4=0$ be $|x|=\alpha ,\beta$.

$\begin{cases} \alpha +\beta = -\dfrac{2a}{4a-15} \\ \alpha \beta = \dfrac{4}{4a-15} \end{cases}$$...(1)$

Important!

WLOG let $\alpha \geq \beta$.

$\alpha \geq 0,\beta \geq 0\implies \alpha +\beta \geq 0,\ \alpha \beta \geq 0$

$\alpha \geq 0,\beta <0\implies \alpha \beta \leq 0$

$\alpha <0,\beta <0\implies \alpha +\beta <0,\alpha \beta >0$

Case I. $\alpha \geq 0,\beta \geq 0$ (Both positive or zero roots.)

$\alpha +\beta \geq 0\implies -\dfrac{2a}{4a-15} \geq 0\implies 2a(4a-15)\leq 0\implies \boxed{0\leq a\leq \dfrac{15}{4}}$

$\alpha \beta \geq 0\implies \dfrac{4}{4a-15} \geq 0\implies 4a-15>0\implies \boxed{a>\dfrac{15}{4}}$

$\dfrac{D}{4} \geq 0\implies a^2-16a+60\geq 0\implies (a-6)(a-10)\geq 0\implies \boxed{a\leq 6,a\geq 10}$

⇒ No solution.

Case II. $\alpha \geq 0,\beta <0$ (One positive or zero root.)

$\alpha \beta \leq 0\implies \dfrac{4}{4a-15} \leq 0\implies 4a-15< 0\implies \boxed{a<\dfrac{15}{4}}$

$\dfrac{D}{4} \geq 0\implies a^2-16a+60\geq 0\implies (a-6)(a-10)\geq 0\implies \boxed{a\leq 6,a\geq 10}$

⇒ The intersection is $\boxed{a< \dfrac{15}{4}}$.

Case III. $\alpha <0,\beta <0$ (Both negative roots.)

$\alpha +\beta <0\implies -\dfrac{2a}{4a-15} <0\implies 2a(4a-15)>0\implies \boxed{a<0,a>\dfrac{15}{4}}$

$\alpha \beta >0\implies \dfrac{4}{4a-15} >0\implies 4a-15>0\implies \boxed{a>\dfrac{15}{4}}$

$\dfrac{D}{4} \geq 0\implies a^2-16a+60\geq 0\implies (a-6)(a-10)\geq 0\implies \boxed{a\leq 6,a\geq 10}$

⇒ The intersection is $\boxed{a>\dfrac{15}{4}}$. This should be rejected as negative numbers cannot be the value of $|x|$.

Case IV. $4a-15=0$ (Linear equation.)

Most of the people get confused here. As the question doesn't state it is a quadratic equation, it can either be a quadratic or linear equation.

Given, $(4a-15)|x|^2+2a|x|+4=0$

$\implies 2a|x|=-4$

$\implies 2\times \dfrac{15}{4} \times |x|=-4$

$\implies |x|=-\dfrac{8}{15}$

⇒ No solution.

Required answer.

$x=\pm\dfrac{-a+ \sqrt{a^2-16a+60} }{4a-15}$ where $a< \dfrac{15}{4}$.

No solution where $a\geq \dfrac{15}{4}$.