Question

Let a, ß are the roots of the equation 2x² - 15x + 4 = 0, then the equation whose roots are a -3,ß- 3 is
a) 2x2 - 27x + 67 = 0
b) 2x2 – 3x – 23 = 0
c) 2x2 + 27x + 67 = 0

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Answer 1

Sol.

Let $\bold{2(x-\alpha )(x-\beta )=2x^2-15x+4}$

By using coefficient comparison method(or, Vieta's formula),

$\begin{cases} & \alpha +\beta =\dfrac{15}{2} \\ & \alpha \beta = 2\end{cases}$

The sum and the product of the new equations are

$\begin{cases} & \alpha +\beta -6=-\dfrac{3}{2} \\ & \alpha \beta-3(\alpha +\beta)+9 = -\dfrac{23}{2} \end{cases}$

And thus the new equation is $\boxed{\bold{2x^2+3x-23=0}}$.

Alternative Method

Let $\bold{f(x)=2x^2-15x+4}$.

Then $\bold{f(x+3)=2(x+3)^2-15(x+3)+4}$.

The required answer is $\boxed{\bold{2x^2-3x-23=0}}$.

What's used

First things first, equation having $\alpha ,\beta$ as zeros is

$2(x-\alpha )(x-\beta )$

by factor theorem.

Here, the new zeros are

$\alpha -3, \beta -3$

The new equation formed is

$2(x-\alpha +3)(x-\beta +3)$

Thus, the equation so formed is

$\boxed{\bold{f(x+3)=0}}$