Question

the roots of the quadratic equation 2x^2+x-1 is​

Answer 1

Step-by-step explanation:

Given equation is 2x

Given equation is 2x 2

Given equation is 2x 2 −x+

Given equation is 2x 2 −x+ 8

Given equation is 2x 2 −x+ 81

Given equation is 2x 2 −x+ 81

Given equation is 2x 2 −x+ 81 =0

Given equation is 2x 2 −x+ 81 =016x

Given equation is 2x 2 −x+ 81 =016x 2

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=0

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=016x

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=016x 2

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=016x 2 −4x−4x+1=0

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=016x 2 −4x−4x+1=04x(4x−1)−1(4x−1)=0

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=016x 2 −4x−4x+1=04x(4x−1)−1(4x−1)=0(4x−1)

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=016x 2 −4x−4x+1=04x(4x−1)−1(4x−1)=0(4x−1) 2

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=016x 2 −4x−4x+1=04x(4x−1)−1(4x−1)=0(4x−1) 2 =0

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=016x 2 −4x−4x+1=04x(4x−1)−1(4x−1)=0(4x−1) 2 =0x=

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=016x 2 −4x−4x+1=04x(4x−1)−1(4x−1)=0(4x−1) 2 =0x= 4

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=016x 2 −4x−4x+1=04x(4x−1)−1(4x−1)=0(4x−1) 2 =0x= 41

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=016x 2 −4x−4x+1=04x(4x−1)−1(4x−1)=0(4x−1) 2 =0x= 41

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=016x 2 −4x−4x+1=04x(4x−1)−1(4x−1)=0(4x−1) 2 =0x= 41 and x=

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=016x 2 −4x−4x+1=04x(4x−1)−1(4x−1)=0(4x−1) 2 =0x= 41 and x= 4

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=016x 2 −4x−4x+1=04x(4x−1)−1(4x−1)=0(4x−1) 2 =0x= 41 and x= 41

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=016x 2 −4x−4x+1=04x(4x−1)−1(4x−1)=0(4x−1) 2 =0x= 41 and x= 41

Given equation is 2x 2 −x+ 81 =016x 2 −8x+1=016x 2 −4x−4x+1=04x(4x−1)−1(4x−1)=0(4x−1) 2 =0x= 41 and x= 41

Answer 2

Answer:

$we \: can \: solve \: by \: spitting \: the \: middle \: term \\ 2x {}^{2} + 2x - x - 1 = 0 \\ (2x + 2x) - (x + 1) = 0\\ 2x( x + 1) - 1(x + 1) = 0 \\ (x + 1)(2x - 1) = 0\\ so \: here \: either \: x + 1 = 0 \: or \: 2x - 1 = 0 \\ x + 1 = 0 \\ x = - 1 \\ 2x - 1 = 0 \\ 2x = 1 \\ x = 1 \div 2 \\ \\ \\ \: roots \: this \: quadratic \: equation \: is \: - 1 \: and \: 1 \div 2$