Question

If α and β are the zeroes of the quadratic polynomial f(x) = 2x
2 – 5x + 7, then find a quadratic
polynomial whose zeroes are 2α + 3β and 2β + 3α .

Answer 1

Answer:

$\boxed{\implies x^2-\frac{25}{2}x+41}$

Step-by-step explanation:

$f(x)=2x^2-5x+7 \\\\ it \ is \ of \ the \ form \ ax^2+bx+c \\ .'. \ a=2,b=-5,c=7\\ \\ \alpha \ and \ \beta \ are \ the \ zeroes \ of \ the \ polynomial. \\\\ => sum \ of \ zeroes = \alpha+\beta \\\\ \alpha+\beta=\frac{-b}{a} = \frac{-(-5)}{2} \\\\ \alpha+\beta=\frac{5}{2} \\\\ =>product \ of \ zeroes=\alpha\beta \\\\ \alpha\beta = \frac{c}{a}=\frac{7}{2} \\ -------------$

$Zeroes \ of \ the \ polynomial \ are \ 2\alpha+3\beta \ and \ 2\beta+3\alpha \\\\ => sum \ of \ zeroes = 2\alpha+3\beta+2\beta+3\alpha \\ \\ = 5\alpha+5\beta \\\\ =5(\alpha+\beta) \\\\ =5(\frac{5}{2}) \\\\ =\frac{25}{2} \\\\ =>product \ of \ zeroes =(2\alpha+3\beta)\times(2\beta+3\alpha) \\\\ =2\alpha(2\beta+3\alpha)+3\beta(2\beta+3\alpha) \\\\ =4\alpha\beta+6\alpha^2+6\beta^2+9\alpha\beta \\\\ =6\alpha^2+6\beta^2+13\alpha\beta \\\\ =6(\alpha^2+\beta^2)+13\alpha\beta$

$= 6[(\alpha+\beta)^2-2\alpha\beta]+13\alpha\beta \\\\ = 6[(\frac{5}{2})^2-2(\frac{7}{2})]+13(\frac{7}{2}) \\\\ =6[\frac{25}{4}-7]+\frac{91}{2} \\\\ =6[\frac{25-28}{4}]+\frac{91}{2} \\\\ =6(\frac{-3}{4})+\frac{91}{2} \\\\=\frac{-9}{2}+\frac{91}{2}\\\\ =\frac{-9+91}{2} \\\\ =\frac{82}{2} \\\\ =41$

$the \ quadratic \ polynomial \ is \ of \ the \ form, \\\implies x^2-[sum \ of \ zeroes]x+[product \ of \ zeroes] \\ \implies x^2-\frac{25}{2}x+41$

hope it helps..!