#### Question Find the value of p, for which one root of the quadratic equation px^{2} – 14x + 8 = 0 is 6 times the other.

#### Video Explanation

#### Explanatory Answer

Let one of the roots be r. Therefore, the second root will be 6r.

The equation is px^{2} – 14x + 8 = 0

In a quadratic equation of the form ax^{2} + bx + c = 0, sum of roots = \\frac{-b}{a})

Sum of the roots r + 6r = 7r = \\frac{-b}{a}) = \\frac{-(-14)}{p})

i.e., 7r = \\frac{14}{p})

Or r = \\frac{2}{p}) ---------------------------(1)

In a quadratic equation of the form ax^{2} + bx + c = 0, product of roots = \\frac{c}{a})

Product of the roots r × 6r = 6r^{2} = \\frac{c}{a}) = \\frac{8}{p}) --------------------(2)

Substitute r = \\frac{2}{p}) in eqn (2)

6\{\left ( {{\frac {2} {p}}} \right )}^{2}) = \\frac{8}{p})

\\frac {24} {{p}^{2}}) = \\frac{8}{p})

Or \\frac{24}{8}) = \\frac {{p}^{2}} {p})

Or p = 3