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The values of 'a' for which quadratic equation (1 – 2a) x2 – 6ax – 1 = 0 and ax2 – x + 1 = 0 have at least one roots in common, are-
- a){0,1/2}
- b){1/2, 2/9}
- c){2/9}
- d){0,1/2,2/9}
Correct answer is option 'C'. Can you explain this answer?
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The values of 'a' for which quadratic equation (1 – 2a) ...
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The values of 'a' for which quadratic equation (1 – 2a) ...
Common Roots of Quadratic Equations
To find the values of \(a\) for which the quadratic equations have at least one root in common, we can use the condition that the discriminant of both equations should be zero.
Quadratic Equation 1
Given equation: \((1 - 2a) x^2 - 6ax - 1 = 0\)
Discriminant of Equation 1
For Equation 1, the discriminant is given by:
\[D_1 = (-6a)^2 - 4(1 - 2a)(-1)\]
\[D_1 = 36a^2 + 4(2a - 1)\]
Quadratic Equation 2
Given equation: \(ax^2 - x + 1 = 0\)
Discriminant of Equation 2
For Equation 2, the discriminant is given by:
\[D_2 = (-1)^2 - 4a(1)\]
\[D_2 = 1 - 4a\]
Setting Discriminants Equal to Zero
For the equations to have at least one root in common, the discriminants should be zero.
So, we have:
\[D_1 = 0 \quad \text{and} \quad D_2 = 0\]
Solving for \(a\)
\[36a^2 + 4(2a - 1) = 0 \implies 36a^2 + 8a - 4 = 0\]
\[D = 64 + 576 = 640\]
As \(D > 0\), the equation has real roots. Therefore, the values of \(a\) for which the equation has at least one root in common are the roots of the equation:
\[a = \frac{-8 \pm \sqrt{640}}{72} = \frac{-8 \pm 16\sqrt{10}}{72} = \frac{-1}{9} \pm \frac{2\sqrt{10}}{9}\]
Hence, the correct answer is option C: \(\frac{2}{9}\).
To find the values of \(a\) for which the quadratic equations have at least one root in common, we can use the condition that the discriminant of both equations should be zero.
Quadratic Equation 1
Given equation: \((1 - 2a) x^2 - 6ax - 1 = 0\)
Discriminant of Equation 1
For Equation 1, the discriminant is given by:
\[D_1 = (-6a)^2 - 4(1 - 2a)(-1)\]
\[D_1 = 36a^2 + 4(2a - 1)\]
Quadratic Equation 2
Given equation: \(ax^2 - x + 1 = 0\)
Discriminant of Equation 2
For Equation 2, the discriminant is given by:
\[D_2 = (-1)^2 - 4a(1)\]
\[D_2 = 1 - 4a\]
Setting Discriminants Equal to Zero
For the equations to have at least one root in common, the discriminants should be zero.
So, we have:
\[D_1 = 0 \quad \text{and} \quad D_2 = 0\]
Solving for \(a\)
\[36a^2 + 4(2a - 1) = 0 \implies 36a^2 + 8a - 4 = 0\]
\[D = 64 + 576 = 640\]
As \(D > 0\), the equation has real roots. Therefore, the values of \(a\) for which the equation has at least one root in common are the roots of the equation:
\[a = \frac{-8 \pm \sqrt{640}}{72} = \frac{-8 \pm 16\sqrt{10}}{72} = \frac{-1}{9} \pm \frac{2\sqrt{10}}{9}\]
Hence, the correct answer is option C: \(\frac{2}{9}\).
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The values of 'a' for which quadratic equation (1 – 2a) ...
Just find out discriminant you will get solution
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The values of 'a' for which quadratic equation (1 – 2a) x2 – 6ax – 1 = 0 and ax2 – x + 1 = 0 have at least one roots in common, are-a){0,1/2}b){1/2, 2/9}c){2/9}d){0,1/2,2/9}Correct answer is option 'C'. Can you explain this answer?
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