Question

3) if the difference of the roots of x^2+2px+q=0 and x^2+2qx+p=0 is equal , then prove that p+q+1=0. ​

Answer 1

Answer:

Given:

The equations are,

x²+2px+q=0

And, x²+2qx+p=0

To prove:

p+q+1=0

Solution:

As we know that,

The equation is of the form,

ax²+bx+c=0

By using quadratic formula,

x=-b±√b²-4ac/2a

x=-b+√b²-4ac/2a,-b-√b²-4ac/2a

Difference of the roots=2√b²-4ac/2a=√b²-4ac/a

→Now,In the given equations,

Difference in the roots are equal,

→√4p²-4q=√4q²-4p

→4p²-4q=4q²-4p

→p²-q=q²-p

→p²-q²+p-q=0

→(p+q)(p-q)+(p-q)=0

→p-q(p+q+1)=0

→p+q+1=0

Step-by-step explanation:

Hope it helps you.......

Answer 2

GIVEN :–

• Difference of the roots of x² + 2px + q = 0 and x² + 2qx + p = 0 is equal.

TO PROVE :–

• p + q + 1 = 0

SOLUTION :–

• If a quadratic equation is ax² + bx + c = 0 then difference of roots is –

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$\bf \large \longrightarrow{ \boxed{ \bf Difference \: \: of \: \:roots = \dfrac{ \pm\sqrt{D} }{a} }}$

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$\bf \large \longrightarrow{ \boxed{ \bf Difference \: \: of \: \:roots = \dfrac{ \pm\sqrt{ {b}^{2} - 4ac} }{a} }}$

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• Difference of roots for x² + 2px + q = 0 is –

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$\bf \implies Difference \: \: of \: \:roots = \dfrac{ \pm\sqrt{ {(2p)}^{2} - 4(1)(q) } }{1}$

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$\bf \implies Difference \: \: of \: \:roots = \pm\sqrt{4 {p}^{2} - 4q} \: \: \: - - - eq.(1)$

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• Difference of roots for x² + 2qx + p = 0 –

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$\bf \implies Difference \: \: of \: \:roots = \dfrac{ \pm\sqrt{ {(2q)}^{2} - 4(1)(p) } }{1}$

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$\bf \implies Difference \: \: of \: \:roots = \pm\sqrt{4 {q}^{2} - 4p} \: \: \: - - - eq.(2)$

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• According to the question –

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$\bf \implies \pm\sqrt{4 {p}^{2} - 4q} = \pm\sqrt{4 {q}^{2} - 4p}$

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• Square on both sides –

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$\bf \implies 4 {p}^{2} - 4q= 4 {q}^{2} - 4p$

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$\bf \implies 4( {p}^{2} - q)= 4 ({q}^{2} - p)$

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$\bf \implies {p}^{2} - q= {q}^{2} - p$

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$\bf \implies {p}^{2} - {q}^{2} = q - p$

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$\bf \implies (p + q) \cancel{(p - q)} = - \cancel{ ( p - q )}$

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$\bf \implies p + q = - 1$

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$\bf \implies \large { \boxed{ \bf p + q + 1 = 0}}$

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$\bf \large \: \: { \underbrace{ \bf \:Hence \: \: proved}}$