KnowledgeBoat Logo

Mathematics

If 23\dfrac{2}{3} and -3 are roots of the equation px2 + 7x + q = 0, find the values of p and q .

Quadratic Equations

46 Likes

Answer

The given equation is px2 + 7x + q = 0.

As 23\dfrac{2}{3} is a root of the equation, so x = 23\dfrac{2}{3} satisfies the given equation.

Substituting x = 23\dfrac{2}{3} in the given equation:

p(23)2+7×23+q=04p9+143+q=0q=4p9143q=(4p+429) …(i)p\big(\dfrac{2}{3} \big)^2 + 7 \times \dfrac{2}{3} + q = 0 \\[1em] \Rightarrow \dfrac{4p}{9} + \dfrac{14}{3} + q = 0 \\[1em] \Rightarrow q = -\dfrac{4p}{9} - \dfrac{14}{3} \\[1em] \Rightarrow q = -\big(\dfrac{4p + 42}{9}\big) \space \text{…(i)} \\[1em]

Also -3 is a root of the given equation, so x = -3 satisfies the given equation.

Substituting x = -3 in the given equation:

p(3)2+7×(3)+q=09p21+q=0 …(ii)p(-3)^2 + 7 \times (-3) + q = 0 \\[0.5em] \Rightarrow 9p - 21 + q = 0 \space \text{…(ii)}

Putting value of q from eqn (i) into eqn (ii)

9p21(4p+429)=09p - 21 - \Big(\dfrac{4p + 42}{9}\Big) = 0

Taking 9 as the LCM

81p1894p429=077p231=077p=231p=23177p=3\Rightarrow \dfrac{81p - 189 - 4p - 42}{9} = 0 \\[1em] \Rightarrow 77p - 231 = 0 \\[1em] \Rightarrow 77p = 231 \\[1em] \Rightarrow p = \dfrac{231}{77} \\[1em] \Rightarrow p = 3

Substituting p = 3 in (i), we get

q=(4×3+429)q=(12+429)q=(549)q=6q = -\Big( \dfrac{4\times 3 + 42}{9} \Big) \\[1em] \Rightarrow q= -\Big(\dfrac{12 + 42}{9}\Big) \\[1em] \Rightarrow q= -\Big(\dfrac{54}{9}\Big) \\[1em] \Rightarrow q= -6

Hence, p = 3 and q = -6

Answered By

33 Likes


Related Questions