Question

Prove that x2+6x+11 has no zero.​

Answer 1

Given a quadratic equation x² + 6x + 11.

To prove that this equation has no real roots.

Solution

Compare it with the standard form of a quadratic equation, ax² + bx + c.

We get a = 1

b = 6

c = 11

Now, if the discriminant (b² - 4ac) is negative or < 0 then the quadratic has no real roots.

So, let's find out the discriminant.

b² - 4ac = 6² - 4(1)(11)

= 36 - 44

= (- 8)

Now, the discriminant is negative, hence the quadratic equation would have no real roots. It would have imaginary roots.

Hence Proved

Answer 2

Solution:

Given equation:

$x {}^{2} + 6x + 11 = 0$

To prove:

The equation doesn't have real roots

Proof:

By completing the square method,

$x {}^{2} + 6x + 11 = 0 \\ \\ \implies \: x {}^{2} + 6x = - 11 \\ \\ \: \implies \: x {}^{2} + 2( \frac{3}{2} )x + ( \frac{3}{2} ) {}^{2} = - 11 + ( \frac{3}{2} ) {}^{2} \\ \\ \implies \: (x + \frac{3}{2} ) {}^{2} = - 11 + \frac{9}{4} \\ \\ \implies \: (x + \frac{3}{2} ) {}^{2} = \frac{ - 35}{4}$

For no value of x,can (x+3/2)^2 have negative value.

As,the condition prevails,we can conclude that the given equation a doesn't have real roots.

Hence,proved