Question

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If the zeroes of the polynomial x2 -px + q are in the ratio 2 : 3, prove that 6p2 = 25q.

Answer 1

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

It is given that zeroes of the polynomial x² - px + q are in the ratio 2 : 3.

Let the zeroes be 2a and 3a.

We know that

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm :\implies\:2a + 3a = - \: \dfrac{( - p)}{1}$

$\rm :\longmapsto\:5a = p$

$\bf\implies \:a = \dfrac{p}{5} - - - (1)$

And

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm :\longmapsto\:3a \times 2a = \dfrac{q}{1}$

$\rm :\longmapsto\: {6a}^{2} = q$

$\rm :\longmapsto\:6 {\bigg(\dfrac{p}{5} \bigg) }^{2} = q$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \bigg \{ \because \:a = \dfrac{p}{5}\bigg \}}$

$\rm :\longmapsto\:6 \times \dfrac{ {p}^{2} }{25} = q$

$\bf\implies \: {6p}^{2} = 25q$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

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