Question

real zeroes of x square-7x+6​

Answer 1

Answer :

x² - 7x + 6

On comparing the given equation with ax² + bx + c = 0, we get,

a = 1, b = - 7, c = 6

D = b² - 4ac

⇒ D = (-7)² - 4 * 1 * 6

⇒ D = 49 - 24

⇒ D = 25 > 0

$\sf x = \frac{ - b \pm \sqrt{D} }{2a} \\ \\ \sf x = \frac{ - ( - 7) \pm \sqrt{25} }{2 \times 1} \\ \\ \sf x = \frac{ 7\pm 5}{2}$

Therefore,

$\bf x = \frac{7+ 5}{2} = \frac{12}{2} = 2 \\ \\ \bf or, \: x = \frac{7 - 5}{2} = \frac{2}{2} = 1$

Hence, the required zeroes are 2 and 1

Answer 2

Solution:-

Given Quadratic Polynomial Equation:-

x² - 7x + 6

To Find :-

Real Roots of the Given Equation = ?

Find:-

x² - 7x + 6

Here,

a = 1

b = -7

c = 6

Since,

It's asked for real Roots, That means real roots exist.

Which mean there will be two Roots and both will be greater than O.

Since, Real Roots exist.

=) Discriminant ( D ) = O

=) b² - 4ac = 0

=) ( -7 )² - 4.1.6

=) 49 - 24

=) 25

Now,

x = [ ( -b ±√D )/ 2a ]

=) x = [ ( 7 ± √25 )/2 ]

=) x = [ ( 7 ± 5 )/2 ]

For x = [ ( 7 + 5 )/2 ]

=) x = 12/2

=) x = 6

For x = [ ( 7 - 5 )/2 ]

=) x = 2/2 = 1

Hence,

The Two Real Roots are 6 and 1.