Question

Question 10
Which of the following equations has no real roots?
(A) $x^2-4x+3\sqrt{2}=0$
(B) $x^2+4x-3\sqrt{2}=0$​​​​​​​
(C) $x^2-4x-3\sqrt{2}=0$​​​​​​​
(D) $3x^2-44\sqrt{3}x+4=0$​​​​​​​

Answer 1

a) The given equation is $x^2-4x+3\sqrt{2}=0$
On comparing with ax2 + bx + c = 0, we get
A = 1, b = -4 and c = $3\sqrt{2}$
The discriminant of x2 – 4x + $3\sqrt{2}$ = 0 is
$D=b^2-4ac$
$=(-4)2-4\left(1\right)\left(3\sqrt{2}\right)=16-12\sqrt{2}=16-12\times \left(1.41\right)$
$=16-16.92=-0.92$
$b^2-4ac<0$
b) iv) Given equation is $3x^2$ – 4x + 1 = 0
on comparing with ax2 + bx + c = 0 we get
a = 1, b = 0 and c = 16
$\therefore discriminateD=b^2-4ac=(-4{)}^2-4(3\left)\right(1)$
= 16 – 12 = 4 > 0 i.e., D > 0
Hence, the equation 3x2 – 4x + 1 = 0 has two distinct real roots
v) Given equation is $(x+4{)}^2$ – 8x = 0
$\Rightarrow x^2+16+8x-8x=0[\therefore (a+b{)}^2=a^2+2ab+b^2]$
$\Rightarrow x^2+16=0$
$\Rightarrow x^2+0x+16=0$
On comparing with $a{x}^2$ + bx + c = 0, we get
a = 1b = 0 and c = 16

discriminant D = $b^2$ – 4ac = $(0{)}^2$ – 4(1) (16) = - 64 < 0 i.e., D<0
Hence, the equation (x + 4)2 – 8x = 0 has imaginary roots, ie., no eral roots
(vi) Given Equation is $(x-\sqrt{2}{)}^2-\sqrt{2}(x+1)=0$
$\Rightarrow x^2+(\sqrt{2}{)}^2-2x\sqrt{2}-\sqrt{2}x-\sqrt{2}=0[\therefore (a-b{)}^2=a^2-2ab+b^2]$
$\Rightarrow x^2-3\sqrt{2}x+(2-\sqrt{2})=0$
On comparing with ax2 + bx + c = 0 we get

a = 1, b = - 3$\sqrt{2}$ and c = 2 – $\sqrt{2}$
$\therefore$ Discriminant. D = $b^2$ – 4ac
$=(-3\sqrt{2})2-4\left(1\right)(2-\sqrt{2})=9\times 2-8+4\sqrt{2}$
=18 – 8 + 4$\sqrt{2}$ = 10 + 4 > 0 i.e., D > 0
Hence, the equation $(x-\sqrt{2})2-\sqrt{2}(x+1)=0$ has two distinct real roots
vii) Given equation is $\frac{3}{\sqrt{2}}x+\frac{1}{\sqrt{2}}=0$
on comparing with $a{x}^2$ + bx + c = 0 we get
$a=\sqrt{2},b=-\frac{3}{\sqrt{2}}andc=\frac{1}{\sqrt{2}}$
Discriminant D = $b^2$ – 4ac
$={(-\frac{3}{\sqrt{2}})}^2-4\sqrt{2}\left(\frac{1}{2}\right)$
$=\frac{9}{2}-4=\frac{9-8}{2}=\frac{1}{2}0i.e.,D>0$
Hence, the equation $\sqrt{2}{x}^2-\frac{3}{\sqrt{2}}x+\frac{1}{\sqrt{2}}=0$ has two distinct real roots
viii) Given equation is x (1 – x) – 2 = 0
$\Rightarrow x-x^2=0$
$\Rightarrow x^2-x+2=0$
On comparing with a $x^2$ + bx + c = 0, we get
A = 1 b = 1 and c = 2
$\therefore$ Discriminant D = $b^2$ – 4ac
= $(-1{)}^2$ – 4( 1)(2) = 1 – 8 = - 7 < 0 i.e., D < 0

Hence the equation x (1 – x) – 2 = 0 has imaginary roots i.e., no real roots
ix) Given equation is
(x – 1) (x + 2) + x = 0
$\Rightarrow x^2+x-2+2=0$
$\Rightarrow x^2+x+0=0$
On comparing the equation with $a{x}^2$ + bx + c – 0, we have
a = 1, b = 1 and c = 0
$\therefore Discriminant,D=b^2-4ac$
= 1 – 4(1)(0) = 1 > 0 i.e., D > 0

Hence, equation has two distinct real roots
x) Given equation is (x + 1) (x – 2) + x = 0
$\Rightarrow x^2+x-2x-2+x=0$
$\Rightarrow x^2-2=0$
$\Rightarrow x^2+0x-2=0$

On comparing with ax2 + bx + c = 0 we get
a = 1, b = 0 and = - 2
$\therefore$ Discriminant, D = $b^2$ -4ac =$(0{)}^2$ – 4 (1) (-2) = 0 + 8 = 8 > 0
Hence the equation (x + 1) (x – 2) + x = 0 has two distinct real roots