Question

One root of the quadratic equation x^2 - 12x + a = 0, is thrice the other. Find the value of a.​

Answer 1

Answer:

hey!⭐

Let the roots of the quadratic equation be x and 3x.

Sum of roots = -(-12) = 12

a + 3a = 4a = 12 => a = 3

Product of the roots = 3a2 = 3(3)2 = 27.

Hope it will be helpful ✌️

Answer 2

Answer:

Required value of a is 27.

Step-by-step explanation:

Given,${x}^{2} - 12x + a = 0....(1)$

is a quadratic equation.

We are comparing equation (1) with $p {x}^{2} + qx + r = 0....(2)$

and get,

$p = 1 \\ q = - 12 \\ r = a$

We know if equation (2) has two roots then,

sum of two roots $= - \frac{q}{p}$

and product of the roots $= \frac{r}{p}$

Let one root of equation (1) be y.

It is also given,one root is thrice of another.

So, other root of equation (2) is 3y.

According to rule,

$y + 3y = - (\frac{ - 12}{1}) \\ 4y = 12 \\ y = \frac{12}{4} \\ y = 3$

and

$y \times 3y = \frac{a}{1} \\ 3 {y}^{2} = a \\ a = 3 \times {3}^{2} \\ a = 3 \times 9 \\ a = 27$

So, required value of a is 27.

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