Question

If 'm' and 'n' are the zeroes of the polynomial 2x^2 + 5x +1 ,then the value of 'm+n' is (a) -2 (b) -1 (c) 1 (d) -5/2 pls answer

Answer 1

Given : -

Quadratic equation

2x² + 5x + 1 = 0

m , n are the zeroes of the polynomial .

Required to find : -

• value of ' m + n ' ?

Formula used : -

Quadratic formula ;

$\boxed{\tt{\bf{ x = \dfrac{ - b \pm \sqrt{ b^2 - 4ac }}{ 2a } }}}$

Solution : -

Quadratic equation ;

2x² + 5x + 1 = 0

m , n are the zeroes of the polynomial .

We need to find the value of " m + n " .

So,

Consider the given quadratic equation .

2x² + 5x + 1 = 0

The standard form of a quadratic equation is

• ax² + bx + c = 0

Comparing the given polynomial with the standard form of polynomial .

Here,

• a = 2

• b = 5

• c = 1

Now,

Using the formula ;

$\boxed{\tt{\bf{ x = \dfrac{ - b \pm \sqrt{ b^2 - 4ac }}{ 2a } }}}$

This implies ;

$\sf \to x = \dfrac{ - \: ( \: 5 \: ) \pm \sqrt{ {5}^{2} - 4(2)(1) } }{2(2)} \\ \\ \to \sf x = \dfrac{ - 5 \pm \sqrt{25 - 8} }{4} \\ \\ \to \sf x = \dfrac{ - 5 \pm \sqrt{17} }{4} \\ \\ \textsf{ This can be written as } \\ \\ \to \sf m = \dfrac{ - 5 + \sqrt{17} }{4} \\ \\ \sf \qquad \qquad \: (and) \: \qquad \qquad \\ \\ \to \sf n = \dfrac{ - 5 - \sqrt{17} }{4}$

Hence,

Value of m = -5+√17/4

Value of n = -5-√17/4

Now,

Let's find the value of " m + n " .

m + n =

$\to \sf \dfrac{ - 5 + \sqrt{17} }{4} + \dfrac{ - 5 - \sqrt{17} }{4} \\ \\ \to \sf \dfrac{ - 5 + \sqrt{17} - 5 - \sqrt{17} }{4} \\ \\ \to \sf \dfrac{ - 5 - 5}{4} \\ \\ \to \sf \dfrac{ - 10}{4} \\ \\ \implies \sf \frac{ - 5}{2}$

Therefore,

Value of m + n = -5/2

Verification

Now,

Let's verify whether our calculations are correct or not .

For this we can use the relationships between the zeroes of the quadratic equation and the coefficients .

So,

The relation between the sum of the zeroes and the coefficients is ;

$\dag{\boxed{\tt{ m + n = \dfrac{ - ( coefficient \ of \ x ) }{ coefficient \ of \ x^2 } }}}$

This implies ;

=> m + n = -b/a

=> m + n = -(5)/2

=> m + n = -5/2

=> -5/2 = -5/2

LHS = RHS

Hence Verified !

Additional Information

The b² - 4ac in the Quadratic formula is known to be as " Discriminate " .

Question :

what is the use of the Quadratic formula ?

Answer :

The Quadratic formula is useful to find the values of the roots of the Polynomial.

The Quadratic formula can be splited as ;

$\boxed{\tt{\bf{ \alpha = \dfrac{ - b \pm \sqrt{ b^2 - 4ac }}{ 2a } }}}$

$\boxed{\tt{\bf{ \beta = \dfrac{ - b \pm \sqrt{ b^2 - 4ac }}{ 2a } }}}$

Here,

alpha , beta are the roots of the Quadratic Equation .

The relationship between the product of the zeroes and the coefficients is ;

$\dag{\boxed{\tt{ m.n = \dfrac{ constant \ term}{ coefficient \ of \ x^2 } }}}$

Here,

m , n are the zeroes of the polynomial .

The number of the roots of a quadratic equation depends upon the degree of the polynomial .

i.e.

If the degree is 1 . It has only 1 root .

If the degree is 2 . It had 2 roots .

If the degree is 3 . It has 3 roots .

If the degree is 4 . It has 4 roots .