Question

If the sum of the squares of the roots of the quadratc equation x^2-8x+k=0 is 40 then find the value of k.
AND
find the sum nd product of roots of 4x^2-2(k+1)x+(k+4)

Answer 1

Answer:

Step-by-step explanation:

1)

Given :

If the sum of the squares of the roots of the quadratic equation $x^2-8x+k=0$ is 40,

then find the value of k.

Solution :

We know that,

A quadratic equation of the form,

$ax^2+bx+c=0$,

The sum of the roots $= \alpha +\beta = \frac{-b}{a}$

Product o the roots $= \alpha \beta =\frac{c}{a}$

In the equation,

$x^2-8x+k=0$

a = 1 ,

b = -8  &

c = k,.

So,

Here,.

⇒ $\alpha +\beta = \frac{-(-8)}{1} = 8$

⇒ $\alpha\beta = k$

Given : $\alpha ^2+\beta ^2 = 40$

__

⇒ $\alpha +\beta = \frac{-(-8)}{1} = 8$

By squaring both the sides,

We get,

⇒ $(\alpha + \beta)^2 = 8^2$

⇒ $\alpha^2 + \beta^2 + 2\alpha \beta = 64$

⇒ $(\alpha^2 + \beta^2) + 2(\alpha \beta) = 64$

⇒ $40 + 2(\alpha \beta) = 64$

⇒ $2(\alpha \beta) = 64-40$

⇒ $2(\alpha \beta) = 24$

⇒ $\alpha \beta= \frac{24}{2}$

⇒ $alpha \beta= 12$ = k

∴ k = 12

__

2)

Given :

To find the sum & product of the roots of the eqution,

$4x^2-2(k+1)x+(k+4)$

Solution :

We know that,

A quadratic equation of the form,

$ax^2+bx+c=0$,

The sum of the roots $= \alpha +\beta = \frac{-b}{a}$

Product o the roots $= \alpha \beta =\frac{c}{a}$

In the equation,

$4x^2-2(k+1)x+(k+4)$

a = 4 ,

b = - 2(k + 1) &

c = (k + 4) ,.

So,

Here,.

⇒ $\alpha +\beta = \frac{-(-2(k+1))}{4} = \frac{2(k+1)}{4} = \frac{1}{2}(k+1)$

∴ The sum of the roots = $\frac{1}{2}(k+1)$

⇒ $\alpha\beta = \frac{(k+4)}{4}$

∴ The product of the roots = $\frac{(k+4)}{4}$