The sum of the roots of the equation, {x | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

${x^2} + |2x - 3| - 4 = 0$

$|2 x-3|=\left\{\begin{array}{ccc}{(2 x-3)} & {	ext { if }} & {x>\frac{3}{2}} \ {-(2 x-3)} & {	ext { i

Vedclass · Accepted Answer

$\sqrt 2$

Vedclass · Answer

${x^2} + |2x - 3| - 4 = 0$ $|2 x-3|=\left\{\begin{array}{ccc}{(2 x-3)} & { ext { if }} & {x>\frac{3}{2}} \ {-(2 x-3)} & { ext { if }} & {x<\frac{3}{2}}\end{array} ight.$ for $x>\frac{3}{2},$ $x^{2}+2 x-3-4=0$ $x^{2}+2 x-7=0$ $x=\frac{-2 \pm \sqrt{4+28}}{2}$ $ = \frac{{ - 2 \pm 4\sqrt 2 }}{2} = - 1 \pm 2\sqrt 2 $ Here $x=2 \sqrt{2}-1$ $\left\{2 \sqrt{2}-1<\frac{3}{2} ight\}$ for $x<\frac{3}{2}$ $x^{2}-2 x+3-4=0$ $\Rightarrow x^{2}-2 x-1=0$ $ \Rightarrow x = \frac{{2 \pm \sqrt {4 + 4} }}{2}$ $= \frac{{2 \pm 2\sqrt 2 }}{2}$ $= 1 \pm \sqrt 2 $ Here $x=1-\sqrt{2} \quad\left\{(1-\sqrt{2})<\frac{3}{2} ight\}$ Sum of roots : $(2 \sqrt{2}-1)+(1-\sqrt{2})=\sqrt{2}$

The sum of the roots of the equation, ${x^2}\, + \,\left| {2x - 3} \right|\, - \,4\, = \,0,$ is

Similar Questions

Let the sum of the maximum and the minimum values of the function $f(x)=\frac{2 x^2-3 x+8}{2 x^2+3 x+8}$ be $\frac{m}{n}$, where $\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$. Then $\mathrm{m}+\mathrm{n}$ is equal to :

The number of solutions of $\frac{{\log 5 + \log ({x^2} + 1)}}{{\log (x - 2)}} = 2$ is

Let $a, b, c, d$ be real numbers between $-5$ and $5$ such that $|a|=\sqrt{4-\sqrt{5-a}},|b|=\sqrt{4+\sqrt{5-b}},|c|=\sqrt{4-\sqrt{5+c}}$ $|d|=\sqrt{4+\sqrt{5+d}}$ Then, the product $a b c d$ is

Let $x, y, z$ be non-zero real numbers such that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=7$ and $\frac{y}{x}+\frac{z}{y}+\frac{x}{z}=9$, then $\frac{x^3}{y^3}+\frac{y^3}{z^3}+\frac{z^3}{x^3}-3$ is equal to

Number of positive integral values of $'K'$ for which the equation $k = \left| {x + \left| {2x - 1} \right|} \right| - \left| {x - \left| {2x - 1} \right|} \right|$ has exactly three real solutions, is