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The number of real values of λ for which the system of linear equations
2x+4y−λz=0
4x+λy+2z=0
λx+2y+2z=0
has infinitely many solutions, is :
2x+4y−λz=0
4x+λy+2z=0
λx+2y+2z=0
has infinitely many solutions, is :
- a)0
- b)1
- c)2
- d)3
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
The number of real values of λ for which the system of linear e...
We are given that $x^2 + 7x + 1 = \sqrt{x^2 + 7x + 1}.$ Squaring both sides, we get
\[(x^2 + 7x + 1)^2 = x^2 + 7x + 1.\]Let $y = x^2 + 7x + 1.$ Then the equation above becomes $y^2 = y.$ Hence, $y(y - 1) = 0,$ which means $x^2 + 7x + 1 = 0$ or $x^2 + 7x = -1.$
The roots of $x^2 + 7x + 1 = 0$ are
\[\frac{-7 \pm \sqrt{45}}{2} = -\frac{7}{2} \pm \frac{\sqrt{45}}{2}.\]The roots of $x^2 + 7x = -1$ are $-\frac{7}{2} \pm \frac{i \sqrt{27}}{2}.$
Thus, there are $\boxed{2}$ real solutions.
\[(x^2 + 7x + 1)^2 = x^2 + 7x + 1.\]Let $y = x^2 + 7x + 1.$ Then the equation above becomes $y^2 = y.$ Hence, $y(y - 1) = 0,$ which means $x^2 + 7x + 1 = 0$ or $x^2 + 7x = -1.$
The roots of $x^2 + 7x + 1 = 0$ are
\[\frac{-7 \pm \sqrt{45}}{2} = -\frac{7}{2} \pm \frac{\sqrt{45}}{2}.\]The roots of $x^2 + 7x = -1$ are $-\frac{7}{2} \pm \frac{i \sqrt{27}}{2}.$
Thus, there are $\boxed{2}$ real solutions.
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The number of real values of λ for which the system of linear equations2x+4y−λz=04x+λy+2z=0λx+2y+2z=0has infinitely many solutions, is :a)0b)1c)2d)3Correct answer is option 'C'. Can you explain this answer?
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