Question

The equation whose roots are 'K' times the roots of the equation ax + bx +c=0 is
1) ax? + Kbx + Kc=0
2) ax?+Kby+K?c=0
3) ax? + Kbx + K?c=0
4) ax +Kºbx+Kc=0​

Answer 1

$\textbf{Given:}$

$\text{Equation is ax^2+bx+c=0 }$

$\textbf{To find:}$

$\text{The equation whose roots are k times roots of the given equation}$

$\textbf{Solution:}$

$\text{Let the roots of ax^2+bx+c=0 be \alpha and \beta }$

$\text{Then,}$

$\alpha+\beta=\dfrac{-b}{a}$

$\alpha\,\beta=\dfrac{c}{a}$

$\text{Clearly, the roots of the required equation are k\,\alpha and k\,\beta }$

$\textbf{Sum of the roots}$

$=k\,\alpha+k\,\beta$

$=k(\alpha+\beta)$

$=k(\dfrac{-b}{a})$

$=\dfrac{-kb}{a}$

$\textbf{Product of the roots}$

$=(k\,\alpha)(k\,\beta)$

$=k^2(\alpha\,\beta)$

$=k^2(\dfrac{c}{a})$

$=\dfrac{k^2c}{a}$

$\textbf{The required equation is}$

$x^2-(\text{sum of the roots})x+(\text{product of the roots})=0$

$x^2-(\dfrac{-kb}{a})x+\dfrac{k^2c}{a}=0$

$x^2+\dfrac{kb}{a}x+\dfrac{k^2c}{a}=0$

$a\,x^2+kb\,x+k^2c=0$

$\therefore\textbf{The required equation is}$

$\boxed{\bf\,a\,x^2+kb\,x+k^2c=0}$

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Answer 2

Given : The equation ax² + bx +c = 0

To find:  The equation whose roots are 'K' times the roots of the equation ax² + bx +c = 0 ?

Solution:

• Now we have given the equation: ax² + bx + c = 0

               x² + bx/a + c/a = 0

• Let the roots be α,β.

               α+β = -b/a and αβ = c/a

• Now multiplying it by k, we get:

               k(α+β) = k(-b/a) and kαkβ = k²(c/a)

• So the equation is:

               x² - (k(α+β))x + k²(αβ) = 0

               x² - (k(-b/a))x + k²(c/a) = 0

               x² + (kb/a)x + k²(c/a) = 0

               ax² + kbx + k²c = 0

Answer:

             So the equation is ax² + kbx + k²c = 0.