Question

Find the nature of the roots of the following quadratic equations:
(i) $2x^2-8x+5=0$

(ii) $3x^2-2\sqrt{6}x+2=0$
(iii) $5x^2-4x+1=0$

(iv) $5x(x-2)+6=0$
(v) $12x^2-4\sqrt{15}x+5=0$

(vi) $x^2-x+2=0$

Answer 1

The given quadratic equations are in the form ofax^2 +bx +c =0\)

We know, determinant $\left(D\right)=b^2–4ac$

$\left(i\right)2x^2-8x+5=0So,a=2,b=-8,c=5Determinant\left(D\right)=b^2–4ac=(-8{)}^2-4(2\left)\right(5)=64–40=24>0$

Since D>0, the determinant of the equation is positive, so the expression does having any real and distinct roots.

$\left(ii\right)3x^2-2\sqrt{6}x+2=0So,a=3b=-2\sqrt{6}c=2\left(D\right)=b^2-4ac=(-2\sqrt{6}{)}^2-4(3\left)\right(2)=24-24=0$

since D=0 two equal real roots

$\left(iii\right)5x^2-4x+1=0a=5b=-4c=1D=b^2-4ac=(-4{)}^2-4(5\left)\right(1)=16-20=-4$

D<0

Since D<0, the determinant of the equation is negative, so the expression does not have any real roots.

$\left(iv\right)5x(x-2)+6=05x^2-10x+6=0a=5b=-10c=6D=b^2-4ac=(-10{)}^2-4(5\left)\right(6)=100-120=-20$
since D<0 there is no real roots

$\left(v\right)12x^2-4\sqrt{15}x+5=0a=12b=-4\sqrt{15}c=5D=b^2-4ac=(-4\sqrt{15}{)}^2-4(12\left)\right(5)=240-240=0$

since D =0
so there are two equal real roots

$\left(vi\right)x^2-x+2=0a=1b=-1c=2D=b^2-4ac=(-1)2-4\left(1\right)\left(2\right)=1-8=-7$

Since D<0
so ther are no real roots