Question

The equation whose roots are smaller by 1 than those of 2x2 – 5x + 6 = 0 is
a) 2x^2– 9x + 13 = 0
b) 2x^2– x + 3 = 0
c) 2x^2+ 9x + 13 = 0
d) 2x^2 + x + 3 = 0​

Answer 1

$\textbf{Given:}$

$2x^2-5x+6=0$

$\textbf{To find:}$

$\text{The equation whose roots are smaller by 1 those of 2x^2-5x+6=0 }$

$\textbf{Solution:}$

$\text{Let the roots of 2x^2-5x+6=0 be \alpha\;\text{and}\;\beta }$

$\text{Then,}$

$\alpha+\beta=\dfrac{-b}{a}=\dfrac{5}{2}$

$\alpha\,\beta=\dfrac{c}{a}=\dfrac{6}{2}=3$

$\text{We have to find the equation whose roots are \alpha-1\;\text{and}\;\beta-1 }$

$\textbf{Sum of the roots}$

$(\alpha-1)+(\beta-1)$

$=(\alpha+\beta)-2$

$=\dfrac{5}{2}-2$

$=\dfrac{5-4}{2}$

$=\dfrac{1}{2}$

$\textbf{Product of the roots}$

$(\alpha-1)(\beta-1)$

$=(\alpha\,\beta)-(\alpha+\beta)+1$

$=3-\dfrac{5}{2}+1$

$=4-\dfrac{5}{2}$

$=\dfrac{8-5}{2}$

$=\dfrac{3}{2}$

$\textbf{The quadratic equation is}$

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

$x^2-\dfrac{1}{2}x+\dfrac{3}{2}=0$

$\implies\bf\,2x^2-x+3=0$

$\therefore\textbf{Option (b) is correct}$

Find more:

If alpha,beta are the roots of equation x^2-5x+6=0 and alpha > beta then the equation with the roots (alpha+beta)and (alpha-beta) is

https://brainly.in/question/5731128

Answer 2

Answer:

2x²-x+3=0

Step-by-step explanation:

WKT, smaller by 1 means f(x+1)=0

ATP,

f(x+1)=2(x+1)²-5(x+1)+6

2(x²+1+2x)-5x-5+6 [(a+b)²=a²+b²-2ab]

2x²+2+4x-5x-5+6

2x²+3-x=2x²-x+3

_HOPE THIS HELPS YOU_