There are 2 quadratic equations, a² –a +p=0 and a² – a + 3p=0 (p≠0). For what value of p will one root of the second equation = double the root of the first equation?

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Solution

The correct option is C -2
Option (c)

Go from answer options

1) Take p=2

a² –a +2=0 and a² – a + 6=0

Roots are = $\frac{1 + \sqrt{(1 + 24)}}{2}$ and $\frac{\sqrt{(1 - 24)}}{2}$

Hence, this option is ruled out

Take p = -2

a² –a -2=0 and a² – a -6=0

Roots of 1^st equation = 2,-1 and roots of second equation = 3, -2

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There are 2 quadratic equations, a2 –a +p=0 and a2 – a + 3p=0 (p≠0). For what value of p will one root of the second equation = double the root of the first equation?

The correct option is C -2Option (c) Go from answer options 1) Take p=2 a2 –a +2=0 and a2 – a + 6=0 Roots are = 1+√(1+24)2 and √(1−24)2 Hence, this option is ruled out Take p = -2 a2 –a -2=0 and a2 – a -6=0 Roots of 1st equation = 2,-1 and roots of second equation = 3, -2

There are 2 quadratic equations, a² –a +p=0 and a² – a + 3p=0 (p≠0). For what value of p will one root of the second equation = double the root of the first equation?