Question

find the nature of root of the equation 2x^2-6x+7​

Answer 1

EXPLANATION.

Quadratic equation.

⇒ 2x² - 6x + 7 = 0.

As we know that,

D = Discriminant  Or b² - 4ac.

⇒ D = (-6)² - 4(2)(7).

⇒ D = 36 - 56.

⇒ D = -20.

⇒ D < 0 Roots are imaginary.

                                                                                                                     

MORE INFORMATION.

Maximum & minimum value of quadratic equation.

In a quadratic expression ax² + bx + c.

(1) = If a > 0, quadratic expression has least value at x = -b/2a. This least value is given by = 4ac - b²/4a = -D/2a.

(2) = If a < 0, quadratic expression has greatest value at x = -b/2a. This greatest value is given by = 4ac - b²/4a = -D/2a

Answer 2

Given Quadratic equation is

$\rm : \implies \: {2x}^{2} - 6x + 7 = 0$

$\blue{\rm : \implies \:On \: comparing \: with \: {ax}^{2} + bx + c= 0, \: we \: get}$

$\rm : \implies \:a \: = \: 2$

$\rm : \implies \:b \: = \: - 6$

$\rm : \implies \:c \: = \: 7$

To find the nature of roots, we have to find Discriminant.

We know,

$\boxed{ \pink{ \bf \: Discriminant \: (D) \: = {b}^{2} - 4ac }}$

Now,

Three cases arises.

• If D > 0, then Quadratic equation have real and distinct roots.

• If D = 0, then Quadratic equation have real and equal roots.

• If D < 0, then Quadratic equation have no real roots.

So,

For the given Quadratic equation

$\rm : \implies \:D \: = \: {( - 6)}^{2} - 4 \times 7 \times 2$

$\rm : \implies \:D \: = 36 - 56$

$\rm : \implies \:D \: = - 20$

$\rm : \implies \:D \: < \: 0$

So, it implies, Quadratic equation has no real roots.