Question

The quadratic equation px2

+ qx – r = 0 cannot have equal roots, if p, q and r > 0. Justify.​

Answer 1

Concept to be used :

★ The possible values of the variable which satisfy the equation are called its roots or solutions .

★ A quadratic equation can have atmost two roots .

★ The general form of a quadratic equation is given as ; ax² + bx + c = 0

★ The discriminant , D of the quadratic equation ax² + bx + c = 0 is given by ;

D = b² - 4ac

★ If D = 0 , then the roots are real and equal .

★ If D > 0 , then the roots are real and distinct .

★ If D < 0 , then the roots are unreal (imaginary) .

Given :

• px² + qx - r = 0

To justify :

• The given quadratic equation can't have equal roots , if p , q , r > 0

Justification :

Here ,

The given quadratic equation is ;

px² + qx - r = 0

Now ,

Comparing the given quadratic equation with the general quadratic equation , ax² + bx + c = 0 , we have ;

a = p

b = q

c = -r

Now ,

The Discriminant of the given quadratic equation will be given as ;

=> D = b² - 4ac

=> D = q² - 4•p•(-r)

=> D = q² + 4pr

Also ,

It is given that ; p , q , r > 0

Thus ,

=> q² + 4pr > 0

=> D > 0

=> D ≠ 0

Clearly ,

The Discriminant of the quadratic equation px² + qx - r = 0 can't be zero if p , q , r > 0 .

Thus ,

The given quadratic equation can't have equal roots (°•° D ≠ 0) .

Hence verified .

Answer 2

Answer:

Solution :-

We know that

D = b² - 4ac

D = Discriminate

b = q

a = p

c = -r

Now,

By putting value

$\large \sf \: D = {q}^{2} - 4(p)( - r)$

$\large \sf \: D = {q}^{2} - 4p \times ( - r)$

$\large \sf \: D = {q}^{2} + 4pr$

Now,

Here,

0 < q² + 4pr

So,

0 < D

Now,

We can clearly see that the discriminant is greater than 0. Hence, proved