Topics
The quadratic equation whose roots are 3 and -1, is
|
A) \( \Large x^{2}-4x+3=0 \) |
|
B) \( \Large x^{2}-2x-3=0 \) |
|
C) \( \Large x^{2}+2x-3=0 \) |
|
D) \( \Large x^{2}+4x+3=0 \) |
Correct Answer:
| B) \( \Large x^{2}-2x-3=0 \) |
Description for Correct answer:
Given that, the roots of the quadratic equation are 3 and -1.
Let \( \Large \alpha = 3 \ and \ \beta = -1 \)
Sum of roots = \( \Large \alpha + \beta = 3 - 1 = 2 \)
Product of roots = \( \Large \alpha . \beta = \left(3\right) \left(-1\right) = -3 \)
Therefore, Required quadratic equation is
\( \Large x^{2} - \left( \alpha + \beta \right) x + \alpha \beta = 0 \)
\( \Large x^{2} - \left(2\right)x + \left(-3\right) = 0 \)
\( \Large x^{2} - 2x - 3 = 0 \)
Part of solved Quadratic Equations questions and answers : >> Elementary Mathematics >> Quadratic Equations
Comments
Similar Questions
1). \( \Large x^{2}+x-20=0; y^{2}-y-30=0 \)
| ||||
2). \( \Large 225x^{2}-4=0; \sqrt{225y}+2=0 \)
| ||||
3). \( \Large \frac{4}{\sqrt{x}}+\frac{7}{\sqrt{x}}=\sqrt{x;} \) \( \Large y^{2}-\frac{ \left(11\right)^{\frac{5}{2}} }{\sqrt{y}} =0 \)
| ||||
4). \( \Large x^{2}-365=364; y-\sqrt{324}=\sqrt{81} \)
| ||||
5). \( \Large 3x^{2}+8x+4=0; 4y^{2}-19y+12=0 \)
| ||||
6). \( \Large x^{2}-x-12=0; y^{2}+5y+6=0 \)
| ||||
7). \( \Large x^{2}-8x+15=0; y^{2}-3y+2=0 \)
| ||||
8). \( \Large x^{2}-32=112; y-\sqrt{169}=0 \)
| ||||
9). \( \Large x-\sqrt{121}=0; y^{2}-121=0 \)
| ||||
10). \( \Large x^{2}-16=0; y^{2}-9y+20=0 \)
|
