Which of the following equations has two distinct real roots?
(a) 2x2-3√2x + 9/4 =0
(b) x2 + x – 5 =0
(c) x2 + 3x + 2√2 =0
(d) 5x2– 3x + 1=0
The given equation is `x^(2)+x-5=0`
On comparing with `ax^(2)+bx+c=0` we get
`a=1, b=1` and `c=-5`
The discriminant of `x^(2)+x-5=0` is
`D=b^(2)-4ax=(1)^(2)-4(1)(-5)`
`=1+20=21`
`implies b^(2)-4acgt0`
So `x^(2)+x-5=0` has two distinct real roots.
(a) Given equation is `2x^(2)-3sqrt(2)x+9//4=0`
On comparing with `ax^(2)+bx+c=0`
`a=2,b=-3sqrt(2)` and `c=9//4`
Now `D=b^(2)-4ac=(-3sqrt(2))^(2)-4(2)(9//4)=18-18=0`
Thus, the equation has real and equal roots.
(c) Given equation is `x^(2)+3x+2sqrt(2)=0`
On comparing with `ax^(2)+bx+c=0`
`a=1,b=3` and `c=2sqrt(2)`
Now `D=b^(2)-4ac=(3)^(2)-4(1)(2sqrt(2))=9-8sqrt(2)lt0`
`:.` Roots of the equation are not real.
(d) Given equation is `5x^(2)+3x+1=0`
On comparing with `ax^(2)+bx+c=0`
`a=5,b=-3, c=1`
Now `D=b^(2)-4ac=(-3)^(2)-4(5)(1)=9-20lt0`
Hence roots of the equation are not real.
(b) x2 + x – 5 = 0
Same as the previous one. Let’s check the discriminant value for distinct real roots.
a. Given, 2x2 – 3√2x + 9/4 = 0
∴ D = b2 – 4ac = 0
= (-3√2)2 – 4(2) (9/4)
= 18 – 18 = 0
Hence, the roots are real and equal.
b. Given, x2 + x – 5 = 0
∴ D = b2 – 4ac = 0
= (1)2 – 4(1) (-5)
= 1 + 20 = 21 > 0
Hence, the roots are real and distinct.
c. Given, x2 + 3x + 2√2 =0
∴ D = b2 – 4ac = 0
= 32 – 4(1) (2√2)
= 9 – 8√2 < 0
Hence, the roots are imaginary.
d. Given, 5x2 – 3x + 1 = 0
∴ D = b2 – 4ac = 0
= (-3)2 – 4(5)(1)
= 9 – 20 < 0
Hence, the roots are imaginary.