Question

solve for x, using quadratic formula p²x²-p²x+q²x-q²=0​

Answer 1

Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

$\rm :\longmapsto\: {p}^{2} {x}^{2} - {p}^{2}x + {q}^{2}x - {q}^{2} = 0$

can be rewritten as

$\rm :\longmapsto\: {p}^{2} {x}^{2} + ( - {p}^{2}+ {q}^{2})x - {q}^{2} = 0$

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

$\boxed{ \tt{ \: a = {p}^{2}}}$

$\boxed{ \tt{ \: b = {q}^{2} - {p}^{2}}}$

$\boxed{ \tt{ \: c = - {q}^{2}}}$

Now,

Discriminant, D of quadratic equation ax² + bx + c = 0 is given by

$\red{\rm :\longmapsto\:D = {b}^{2} - 4ac}$

On Substituting the values of a, b and c, we get

$\rm \: = \: {( {q}^{2} - {p}^{2}) }^{2} - 4 \times {p}^{2} \times ( - {q}^{2})$

$\rm \: = \: {( {q}^{2} - {p}^{2}) }^{2} + 4 {q}^{2} {p}^{2}$

We know,

$\boxed{ \tt{ \: {(x - y)}^{2} + 4xy = {(x + y)}^{2} \: }}$

So, using this, we get

$\rm \: = \: {( {q}^{2} + {p}^{2} )}^{2}$

$\purple{\bf\implies \: \boxed{ \tt{ \:D \: = \: {( {q}^{2} + {p}^{2} )}^{2}}}}$

Now, we know, Quadratic formula is

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

or

$\rm :\longmapsto\:\boxed{ \tt{ \: x = \dfrac{ - b \: \pm \: \sqrt{ D}}{2a} \: \: }}$

So, on substituting the values, we get

$\rm :\longmapsto\:x = \dfrac{ - ( {q}^{2} - {p}^{2}) \: \pm \: \sqrt{ {( {q}^{2} + {p}^{2})}^{2} } }{2 {p}^{2} }$

$\rm :\longmapsto\:x = \dfrac{ {p}^{2} - {q}^{2} \: \pm \: ( {q}^{2} + {p}^{2})}{ {2p}^{2} }$

$\rm :\longmapsto\:x = \dfrac{ {p}^{2} - {q}^{2} + ( {q}^{2} + {p}^{2})}{ {2p}^{2} } \: \: or \: \: \dfrac{ {p}^{2} - {q}^{2} - ( {q}^{2} + {p}^{2})}{ {2p}^{2} }$

$\rm :\longmapsto\:x = \dfrac{ {p}^{2} - {q}^{2}+{q}^{2} +{p}^{2}}{ {2p}^{2} } \: \: or \: \: \dfrac{ {p}^{2} -{q}^{2} -{q}^{2} - {p}^{2}}{ {2p}^{2} }$

$\rm :\longmapsto\:x = \dfrac{ 2{p}^{2}}{ {2p}^{2} } \: \: or \: \: \dfrac{ -2{q}^{2} }{ {2p}^{2} }$

$\rm :\longmapsto\:x = 1 \: \: or \: \: \dfrac{ -{q}^{2} }{ {p}^{2} }$

$\bf\implies \:\boxed{ \tt{ \: x = 1 \: \: or \: \: x = - \: \frac{ {q}^{2} }{ {p}^{2} } \: \: }}$

More to know :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac