Question

If $\alpha ,\beta$ are the roots of the equation $2x^2-5x+7=0,$ the equation whose roots are $(2\alpha +3\beta )$ and $(3\alpha +2\beta )$ is:

• A. $2x^2+25x-82=0$
• B. $2x^2-25x+82=0$
• C. $2x^2+25x+82=0$
• D. None of these

Answer 1

The correct option is B $2x^2-25x+82=0$

The correct option is B

$p\left(x\right)=2x²-5x+7$

$a=2,b=-5,c=7$

α and β are the zeros of p(x)

we know that,

sum of zeros = $\alpha +\beta$

$=\frac{-b}{a}$

$=\frac{5}{2}$

product of zeros = $\frac{c}{a}$

$=\frac{7}{2}$

$2\alpha +3\beta$ and $3\alpha +2\beta$ are zeros of a polynomial.

sum of zeros

$=2\alpha +3\beta +3\alpha +2\beta$

$=5\alpha +5\beta$

$=5[\alpha +\beta ]$

$=5\times 5/2$

$=25/2$

product of zeros = $(2\alpha +3\beta )(3\alpha +2\beta )$

$=2\alpha [3\alpha +2\beta ]+3\beta [3\alpha +2\beta ]$

$=6\alpha ²+4\alpha \beta +9\alpha \beta +6\beta ²$

$=6\alpha ²+13\alpha \beta +6\beta ²$

$=6[\alpha ²+\beta ²]+13\alpha \beta$

$=6\left[\right(\alpha +\beta )²-2\alpha \beta ]+13\alpha \beta$

$=6\left[\right(5/2)²-2\times 7/2]+13\times 7/2$

$=6[25/4-7]+91/2$

$=6[25/4-28/4]+91/2$

$=6[-3/4]+91/2$

$=-18/4+91/2$

$=-9/2+91/2$

$=82/2$

$=41$

$\frac{-18}{4}=\frac{-9}{2}$ [ simplest form ]

a quadratic polynomial is given by :-
$k\left(x²-\left(sumofzeros\right)x+\left(productofzeros\right)\right)$
$k\left(x²-\frac{5}{2}x+41\right)$
$k=2$
$=2(x²-\frac{25}{2}x+41)$
$=2x²-25x+82$