Question

Which of the following equation ha no real root.
a› x²+ 4x - 3 √ 2 = 0
b› x²- 4x + 3 √ 2 = 0
c›x²-4x - 3 √ 2 = 0
d›x²+ 4x + 3 √ 2 = 0 ​

Answer 1

Question :–

● Which of the following equation has no real root.

(a) x² + 4x - 3√2 = 0

(b) x² - 4x + 3√2 = 0

(c) x² - 4x - 3√2 = 0

(d) x² + 4x + 3√2 = 0

ANSWER :–

• To know the nature of roots , we have to calculate Discriminant of the quadratic equation ax² + bx + c = 0.

$\\ \bigstar \: \large \: \: { \red{ \boxed{ \bold{D = {b}^{2} - 4ac }}}}$

☞ When –

$\\ \: \: \: \: \:{ \huge{.}} \: \: \: \: { \blue{ \bold{D > 0 , \: \: roots \: \: are \: \: real \: \: and \: \: Distinct.}}}$

$\\ \: \: \: \: \:{ \huge{.}} \: \: \: \: { \blue{ \bold{D = 0 \: \: , \: \: roots \: \: are \: \: real \: \: and \: \: same.}}}$

$\\ \: \: \: \: \:{ \huge{.}} \: \: \: \: { \blue{ \bold{D < 0 \: \: , \: \: roots \: \: are \: \: imaginary. }}}$

OPTION (a) :–

• x² + 4x - 3√2 = 0

$\\ \implies \: { \bold{D = {b}^{2} - 4ac }}$

$\\ \implies \: { \bold{D = {(4)}^{2} - 4(1)( - 3 \sqrt{2}) }}$

$\\ \implies \: { \bold{D = 16- 4( - 3 \sqrt{2}) }}$

$\\ \implies \: { \bold{D = 16 + 12\sqrt{2} > 0 }}$

$\\ \longrightarrow \: { \bold{ Hence \: \: roots \: \: are \: \: real \: \: and \: \: Distinct . }}$

OPTION (b) :–

• x² - 4x + 3√2 = 0

$\\ \implies \: { \bold{D = {b}^{2} - 4ac }}$

$\\ \implies \: { \bold{D = {( - 4)}^{2} - 4(1)( 3 \sqrt{2}) }}$

$\\ \implies \: { \bold{D = 16- 4( 3 \sqrt{2}) }}$

$\\ \implies \: { \bold{D = 16 - 12\sqrt{2} }}$

$\\ \implies \: { \bold{D = - 0.97 (Approximate) }}$

$\\ \implies \: { \bold{D < 0}}$

$\\ \longrightarrow \: { \bold{ Hence \: \: roots \: \: are \: \: imaginary. }}$

OPTION (c) :–

• x² - 4x - 3√2 = 0

$\\ \implies \: { \bold{D = {b}^{2} - 4ac }}$

$\\ \implies \: { \bold{D = {( - 4)}^{2} - 4(1)( - 3 \sqrt{2}) }}$

$\\ \implies \: { \bold{D = 16- 4( - 3 \sqrt{2}) }}$

$\\ \implies \: { \bold{D = 16 + 12\sqrt{2} > 0 }}$

$\\ \longrightarrow \: { \bold{ Hence \: \: roots \: \: are \: \: real \: \: and \: \: Distinct . }}$

OPTION (d) :–

• x² + 4x + 3√2 = 0

$\\ \implies \: { \bold{D = {b}^{2} - 4ac }}$

$\\ \implies \: { \bold{D = {( 4)}^{2} - 4(1)( 3 \sqrt{2}) }}$

$\\ \implies \: { \bold{D = 16- 4( 3 \sqrt{2}) }}$

$\\ \implies \: { \bold{D = 16 - 12\sqrt{2} }}$

$\\ \implies \: { \bold{D = - 0.97 (Approximate) }}$

$\\ \implies \: { \bold{D < 0}}$

$\\ \longrightarrow \: { \bold{ Hence \: \: roots \: \: are \: \: imaginary. }}$

• So that , OPTION (b) & OPTION (d) have no real roots.

• Hence , OPTION (b) & (d) are correct.

Answer 2

Step-by-step explanation:

The quadratic equation ax

2

+bx+c=0 has no real roots only when discriminant D<0

D=b

2

−4ac<0

A) x

2

+4x+3

2

=0

Here,

D=(4)

2

−4×3

2

=16−16.97<0

B) x

2

+4x−3

2

=0

Here,

D=(4)

2

−4×(−3

2

)

=16+16.97>0

C) x

2

+5x+3

2

=0

Here,

D=(5)

2

−4×(3

2

)

=25−16.97>0

D) 3x

2

+4

3

x+4=0

Here,

D=(4

3

)

2

−4×3×4

=48−48=0