Question

If one root of the quadratic equation ax2 +bx+c=0 is triple the other show that 3b2 = 16 ac

Answer 1

Let the roots be alpha and 3alpha

=> alpha+3alpha = -b/a
=> 4alpha = -b/a
=> alpha = -b/4a ------(1)
=> alpha*3alpha = c/a
=> 3(alpha)^2 = c/a -----(2)

substitute (1) in (2)

=> 3(-b/4a)^2 = c/a
=> b^2/16a^2 = c/3a
=> b^2 = (16a^2*c)/3a
=> b^2 = 16ac/3
=> 3b^2 = 16ac (hence proved)

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Answer 2

Step-by-step explanation:

$as \: per \: condition \\ \alpha = 3 \beta \\ thereafter. \: \\ \alpha + 3 \alpha = \frac{ - b}{a} \: \: (bcz. \alpha + \beta = \frac{ - b}{a} ) \\ 4 \alpha = \frac{ - b}{a} \\ \alpha = \frac{ - b}{4 \alpha } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - (1) \\ now \\ \alpha \times 3 \alpha = \frac{c}{a} (bcz.\alpha \times \beta = \frac{c}{a}) \\ 3 { \alpha }^{2} = \frac{c}{a} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - (2) \\ substituting \: the \: value \: of \: \alpha \: from \: 1st \\ therefore \\ 3 \times { (\frac{ - b}{4a}) }^{2} = \frac{c}{a} \\ \frac{3 {b}^{2} }{16 {a}^{2} } = \: \frac{c}{a} \\ 3 {b}^{2} = \frac{16 {a}^{2}c }{a} \\ 3 {b}^{2} = 16ac$

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