Question:
A value of $b$ for which the equations $x^2+bx-1=0,x^2+x+b=0$ have one root in common is
A value of $b$ for which the equations $x^2+bx-1=0,x^2+x+b=0$ have one root in common is
Updated On: Aug 31, 2023
- $-\sqrt2$
- $-i\sqrt3$
- $i\sqrt5$
- $\sqrt2$
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The Correct Option is B
Solution and Explanation
The correct answer is B:\(-i\sqrt3\)
If \(a_1x^2+b_1x+c_1=0 \ and\ a_2x^2+b_2x+c_2=0\)
have a common real root, then
\(\Rightarrow\space20mm(a_1 c_2-a_2 c_1)^2=(b_1 c_2-b_2 c_1)(a_1 b_2-a_2 b_1)\) \[\begin{array}{l} x^2 + bx - 1 = 0 \\ x^2 + x + b = 0 \end{array} \quad \Bigg\{ \text{have a common root}\]\(\Rightarrow\space25mm (1+b^2)=(b^2+1)(1-b)\)
\(\Rightarrow\space20mmb^2+2b+1=b^2-b^3+1-b\)
\(\Rightarrow\space25mm b^3+3b=0\)
\(\therefore\space25mm b(b^2+3)=0 \Rightarrow b=0,\pm \sqrt3 \ i\)
If \(a_1x^2+b_1x+c_1=0 \ and\ a_2x^2+b_2x+c_2=0\)
have a common real root, then
\(\Rightarrow\space20mm(a_1 c_2-a_2 c_1)^2=(b_1 c_2-b_2 c_1)(a_1 b_2-a_2 b_1)\) \[\begin{array}{l} x^2 + bx - 1 = 0 \\ x^2 + x + b = 0 \end{array} \quad \Bigg\{ \text{have a common root}\]\(\Rightarrow\space25mm (1+b^2)=(b^2+1)(1-b)\)
\(\Rightarrow\space20mmb^2+2b+1=b^2-b^3+1-b\)
\(\Rightarrow\space25mm b^3+3b=0\)
\(\therefore\space25mm b(b^2+3)=0 \Rightarrow b=0,\pm \sqrt3 \ i\)
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
