The equation whose roots are obtained by adding 1 to those of 2x2 + 3x

1.	The equation whose roots are obtained by adding 1 to those of 2x2 + 3x + 5 = 0 is A) 2x2 – x – 4 = 0 B) 2x2 + x – 4 = 0 C) 2x2 – x + 4 = 0 D) None
Answer» Correct option is (C) 2x² – x + 4 = 0 Let \(\alpha\;and\;\beta\) are root of \(2x^2+3x+5=0\) \(\therefore\) \(\alpha+\beta\) \(=\frac{-3}2\) and \(\alpha\beta\) \(=\frac{5}2\) Now, \((\alpha+1)+(\beta+1)\) \(=\alpha+\beta+2\) \(=\frac{-3}2+2\) \(=\frac{-3+4}2=\frac12\) and \((\alpha+1)(\beta+1)\) \(=\alpha\beta+\alpha+\beta+1\) \(=\frac52-\frac32+1\) = 1+1 = 2 \(\therefore\) Equation whose roots are \((\alpha+1)\;and\;(\beta+1)\) is \(x^2-((\alpha+1)+(\beta+1))x+(\alpha+1)(\beta+1)=0\) \(\Rightarrow\) \(x^2-\frac x2+2=0\) \(\Rightarrow\) \(2x^2-x+4=0\) \(\therefore\) Required equation is \(2x^2-x+4=0.\) Correct option is C) 2x² – x + 4 = 0

The equation whose roots are obtained by adding 1 to those of 2x2 + 3x + 5 = 0 is A) 2x2 – x – 4 = 0 B) 2x2 + x – 4 = 0 C) 2x2 – x + 4 = 0 D) None

Answer»

Correct option is (C) 2x² – x + 4 = 0

Let \(\alpha\;and\;\beta\) are root of \(2x^2+3x+5=0\)

\(\therefore\) \(\alpha+\beta\) \(=\frac{-3}2\) and \(\alpha\beta\) \(=\frac{5}2\)

Now, \((\alpha+1)+(\beta+1)\) \(=\alpha+\beta+2\) \(=\frac{-3}2+2\) \(=\frac{-3+4}2=\frac12\)

and \((\alpha+1)(\beta+1)\) \(=\alpha\beta+\alpha+\beta+1\) \(=\frac52-\frac32+1\) = 1+1 = 2

\(\therefore\) Equation whose roots are \((\alpha+1)\;and\;(\beta+1)\) is \(x^2-((\alpha+1)+(\beta+1))x+(\alpha+1)(\beta+1)=0\)

\(\Rightarrow\) \(x^2-\frac x2+2=0\)

\(\Rightarrow\) \(2x^2-x+4=0\)

\(\therefore\) Required equation is \(2x^2-x+4=0.\)

Correct option is C) 2x² – x + 4 = 0