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Question: Show that the roots of the equation \({x^2} - 2x + 3 = 0\) are imaginary....

Show that the roots of the equation x22x+3=0{x^2} - 2x + 3 = 0 are imaginary.

Explanation

Solution

The given equation is a quadratic equation so it will have two roots. The roots of this equation can be found out by using the formula method. The formula method states that, the roots of a quadratic equation of the form ax2+bx+c=0a{x^2} + bx + c = 0 is given by x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} where b24ac=Δ{b^2} - 4ac = \Delta

Complete step-by-step solution:
The given quadratic equation is x22x+3=0{x^2} - 2x + 3 = 0.
By comparing it with the standard quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0, we get a=1,b=2a = 1,b = - 2 and c=3c = 3.
Substituting these values in b24ac=Δ{b^2} - 4ac = \Delta ,
Δ=(2)24(1)(3)\Delta = {( - 2)^2} - 4(1)(3)
Solving the square term and subtracting,
Δ=412=8\Delta = 4 - 12 = - 8
Δ=8<0\therefore \Delta = - 8 < 0
Since b24ac<0{b^2} - 4ac < 0, we can say that the roots of this equation are imaginary.
To find the roots we use the formula x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} and substitute all the values,
x=(2)±82(1)x = \dfrac{{ - ( - 2) \pm \sqrt { - 8} }}{{2(1)}}
Solving the brackets,
x=2±82x = \dfrac{{2 \pm \sqrt { - 8} }}{2}
We can write 8=(1)(8) - 8 = ( - 1)(8),
x=2±(1)(8)2x = \dfrac{{2 \pm \sqrt {( - 1)(8)} }}{2}
We know the value of 1=i\sqrt { - 1} = i,
x=2±i82\therefore x = \dfrac{{2 \pm i\sqrt 8 }}{2}
Simplifying further,
x=2±i2×2×22x = \dfrac{{2 \pm i\sqrt {2 \times 2 \times 2} }}{2}
Taking a pair of 22outside the root,
x=2±2i22x = \dfrac{{2 \pm 2i\sqrt 2 }}{2}
Taking 22 common from the numerator,
x=2(1±i2)2x = \dfrac{{2(1 \pm i\sqrt 2 )}}{2}
Cancelling 22from numerator and denominator,
x=1±i2x = 1 \pm i\sqrt 2
Therefore, we get the roots of x22x+3=0{x^2} - 2x + 3 = 0 as 1+i21 + i\sqrt 2 and 1i21 - i\sqrt 2 and both the roots are imaginary.
Hence, we found that the roots of x22x+3=0{x^2} - 2x + 3 = 0 are imaginary.

Note: In the equation ax2+bx+c=0a{x^2} + bx + c = 0if b24ac<0{b^2} - 4ac < 0, then the roots of the quadratic equation are imaginary, if b24ac>0{b^2} - 4ac > 0, then the roots of the quadratic equation are real and unequal and if b24ac=0{b^2} - 4ac = 0, then the roots are real and equal of the quadratic equation.