Question

The value of a for which one root of the equation x2 - 20x- a = 0 is the square of other root, is/are

Answer 1

Given,

A quadratic equation x²-20x-a = 0.

One root of the equation is the square of the other.

To Find,

The value of a.

Solution,

Let the roots of the equation be c and c².

The given equation is x²-20x-a = 0.

Sum of roots = -b/a

where, b is the coefficient of x and a is the coefficient of x².

So,

c+c² = 20

c(c+1)  = 20

Comparing this equation with 4*5 = 20

So, c = 4

Product of roots = c/a

c³ = -a

4³ = -a

a = -64

Hence, the value of a is -64.

Answer 2

Given equation is

${x}^{2} - 20x - a = 0$

It is a quadratic equation and we get two roots for a quadratic equation.

Let the roots of this equation be

$\alpha \\ \beta$

Given that one root is a square of the other root.

Let

$\beta = { \alpha }^{2}$

Now, we know that sum of the roots is

$\alpha + \beta = - b \div a$

And product of the roots is

$\alpha \times \beta = c \div a$

Let us compare the given equation with the standard quadratic form

$a {x}^{2} + bx + c = 0$

From the given equation

So,

$a = 1 \\ b = - 20 \\ c = - a$

By substituting these values in the sum of the roots relation we get,

$\alpha + \beta = 20 \\ \alpha + { \alpha }^{2} = 20$

Let

$\alpha = 4 \\ { \alpha }^{2} = 16$

So,

$\alpha + { \alpha }^{2} \\ = 4 + 16 \\ = 20$

Therefore, it's ture and the value of

$\alpha = 4$

Now, from the relation of product of roots

$\alpha \times \beta = - a \\ \alpha \times { \alpha }^{2} = - a \\ { \alpha }^{3} = - a \\ {4}^{3} = - a \\ - a = 64 \\ a = - 64$

Hence, the value of a is $a = - 64$

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