**Question:**

Which of the following equations has no real roots?

(a) x2 – 4x + 3√2 =0

(b)x2+4x-3√2=0

(c) x2 – 4x – 3√2 = 0

(d) 3x2 + 4√3x + 4=0

**Solution:**

**(a)**

(a) The given equation is $x^{2}-4 x+3 \sqrt{2}=0$.

On comparing with $a x^{2}+b x+c=0$, we get

$a=1, b=-4$ and $c=3 \sqrt{2}$

The discriminant of $x^{2}-4 x+3 \sqrt{2}=0$ is

$D=b^{2}-4 a c$

$=(-4)^{2}-4(1)(3 \sqrt{2})=16-12 \sqrt{2}=16-12 \times(1.41)$

$=16-16.92=-0.92$

$\Rightarrow \quad b^{2}-4 a c<0$

(b) Tho givan onuation is $x^{2}+4 x-3 \sqrt{2}=0$

On comparing the equation with $a x^{2}+b x+c=0$, we get

$a=1, b=4$ and $c=-3 \sqrt{2}$

Then, $D=b^{2}-4 a c=(-4)^{2}-4(1)(-3 \sqrt{2})$

$=16+12 \sqrt{2}>0$

Hence, the equation has real roots.

(c) Given equation is $x^{2}-4 x-3 \sqrt{2}=0$

On comparing the equation with $a x^{2}+b x+c=0$, we get

$a=1, b=-4$ and $c=-3 \sqrt{2}$

Then, $D=b^{2}-4 a c=(-4)^{2}-4(1)(-3 \sqrt{2})$

$=16+12 \sqrt{2}>0$

Hence, the equation has real roots.

(d) Given equation is $3 x^{2}+4 \sqrt{3} x+4=0$.

On comparing the equation with $a x^{2}+b x+c=0$, we get

$a=3, b=4 \sqrt{3}$ and $c=4$

Then, $\quad D=b^{2}-4 a c=(4 \sqrt{3})^{2}-4(3)(4)=48-48=0$

Hence, the equation has real roots.

Hence, $x^{2}-4 x+3 \sqrt{2}=0$ has no real roots.