Question

2x² - √3x + 2 = 0 Find the two roots of quadratic equation.​

Answer 1

Step-by-step explanation:

Given equation is 2x² - √3x + 2 = 0

=>2x²-√3x+2=0

Dividing by 2

=>(2x²/2)-(√3/2)x+(2/2)=(0/2)

=>x²-√3/2 x+1=0

=>x²-(√3/2)x=-1

=>x²-(2/2)(√3/2)x=-1

=>x²-(2)(x)(√3/4)=-1

adding (√3/4)² on both sides

=>x²-2(x)(√3/4)+(√3/4)²=-1+(√3/4)²

=>(x-√3/4)²=-1+(9/16)

=>(x-√3/4)²=(-16+9)/16

=>(x-√3/4)²=-7/16

=>(x-√3/4)=±√(-7/16)

=>x=(√3/4)±√-7/4

=>x=(√3±√-7)/4

Roots are (√3+√-7)/4 and (√3-√-7)/4

Using method:-

Completing the square method

Answer 2

Step-by-step explanation:

Given equation is 2x² - √3x + 2 = 0

=>2x²-√3x+2=0

Dividing by 2

=>(2x²/2)-(√3/2)x+(2/2)=(0/2)

=>x²-√3/2 x+1=0

=>x²-(√3/2)x=-1

=>x²-(2/2)(√3/2)x=-1

=>x²-(2)(x)(√3/4)=-1

adding (√3/4)² on both sides

=>x²-2(x)(√3/4)+(√3/4)²=-1+(√3/4)²

=>(x-√3/4)²=-1+(9/16)

=>(x-√3/4)²=(-16+9)/16

=>(x-√3/4)²=-7/16

=>(x-√3/4)=±√(-7/16)

=>x=(√3/4)±√-7/4

=>x=(√3±√-7)/4

Roots are (√3+√-7)/4 and (√3-√-7)/4

Using method:-

Completing the square method