# How do you find the domain of f(x)=sqrt((2x^2+5x-3) )?

Jan 5, 2018

${D}_{f} = \left(- \infty , - 3\right] \cup \left[\frac{1}{2} , + \infty\right)$

#### Explanation:

$f \left(x\right) = \sqrt{2 {x}^{2} + 5 x - 3}$

D_f={AAx$\in$$\mathbb{R}$: 2x^2+5x-3>=0}

Let's find the roots of $2 {x}^{2} + 5 x - 3 = 0$

Δ=b^2-4ac=25-4*2*(-3)=25+24=49

x_(1,2)=(-b+-sqrtΔ)/(2a) $=$ $\frac{- 5 \pm \sqrt{49}}{4}$ $=$ $\frac{- 5 \pm 7}{4}$

So

• ${x}_{1} = \frac{2}{4} = \frac{1}{2}$,
• ${x}_{2} = - \frac{12}{4} = - 3$

For $x$$\in$$\left(- \infty , - 3\right)$
for example $x = - 5$ $\to$ $2 \cdot {\left(- 5\right)}^{2} + 5 \left(- 5\right) - 3 = 50 - 25 - 3 = 22 > 0$

For $x$$\in$$\left(- 3 , \frac{1}{2}\right)$
for example $x = 0$ $\to$ $2 \cdot {0}^{2} + 5 \cdot 0 - 3 = - 3 < 0$

For $x$$\in$$\left(\frac{1}{2} , + \infty\right)$
for example $x = 5$ $\to$ $2 \cdot {\left(5\right)}^{2} + 5 \cdot 5 - 3 = 50 + 25 - 3 = 72 > 0$

Therefore $2 {x}^{2} + 5 x - 3 \ge 0$ $\iff$ $x$$\in$$\left(- \infty , - 3\right] \cup \left[\frac{1}{2} , + \infty\right)$

and ${D}_{f} = \left(- \infty , - 3\right] \cup \left[\frac{1}{2} , + \infty\right)$