Question

find the roots of quadric equation 2x square +x-4=o using quadric formula​

Answer 1

Solution:-

Given equation is

$\rm \to2 {x}^{2} + x - 4 = 0$

Compare with

$\rm \to \: a {x}^{2} + bx + c = 0$

So

$\to \rm \: a \: = 2,b = 1 \: and \: c = - 4$

Quadratic formula is

$\rm \: x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

Put the value on formula

$\rm \to \: x = \dfrac{ - 1 \pm \sqrt{ {1}^{2} - 4 \times 2 \times - 4} }{2 \times 2}$

$\to\rm \: x = \dfrac{ - 1 \pm \sqrt{1 + 32} }{4}$

$\to\rm \: x = \dfrac{ - 1 \pm \sqrt{33} }{4}$

$\rm \to \: x = \dfrac{ - 1 + \sqrt{33} }{4} ,x = \dfrac{ - 1 - \sqrt{33} }{4}$

Answer

$\rm \to \: x = \dfrac{ - 1 + \sqrt{33} }{4} ,x = \dfrac{ - 1 - \sqrt{33} }{4}$

Answer 2

Step-by-step explanation:

Given:-

${2x}^{2 }+x-4=0$

To find:-

The roots of equation

Solution:-

Here

by comparing with equation

${ax}^{2}+{bx}+c$ we get

a=2,b=1,c=-4

Let's the roots are=${\alpha} \:and\:{\beta}$

Let's put the quadric formula

${\boxed{{\alpha}={\frac {-b+{\sqrt {{b}^{2}-4ac}}}{2a}}}}$

[substitute the values]

${:}\longrightarrow$${\cfrac {(-1)+{\sqrt {{(1)}^{2}-4×2×(-4)}}}{2×2}}$

${:}\longrightarrow$${\dfrac{(-1)+{\sqrt {1+32}}}{4}}$

${:}\longrightarrow$${\underline{\boxed{\bf {{\alpha}={\dfrac{-1+{\sqrt {33}}}{4}}}}}}$

${\boxed{{\beta}={\dfrac {-b-{\sqrt {{b}^{2}-4ac}}}{2a}}}}$

[Substitute the values]

${:}\longrightarrow$${\dfrac {(-1)-{\sqrt {{(1)}^{2}-4×2×(-4)}}}{2×2}}$

${:}\longrightarrow$${\dfrac {(-1)-{\sqrt {1+32}}}{4}}$

${:}\longrightarrow$${\underline{\boxed{\bf {{\beta}={\dfrac {-1-{\sqrt{33}}}{4}}}}}}$

$\therefore$The roots of equation are

${\underline{\boxed{\bf {{\dfrac{-1+{\sqrt {33}}}{4}}\:,\:{\dfrac {-1-{\sqrt {33}}}{4 }}}}}}$

Extra information:-

• The Quadric formula is

${\boxed {x={\cfrac {-b{\underline {+}}{\sqrt {{b}^{2}-4ac}}}{2a}}}}$