Quadratic Equation
Direction: In each of the following question, there are two equations. you have to solve both equations and give answer.
- 225x2 - 4 = 0; √225y + 2 = 0
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Solve both equation and find the value of x and y.
Correct Option: E
225x2 - 4 = 0;
⇒ 225x2 = 4 ⇒ x2 = 4/225
∴ x = √4/225 = ± 2/15, i.e., 2/15 and -2/15
and √225y + 2 = 0 or √225y = -2
On squaring both sides, we get
(√225y)2 = (-2)2
⇒ 225y = 4
∴ y = 4/225
So, relation cannot be established because 4/225 lies between 2/15 and -2/15.
- If one of the roots of quadratic equation 7y2 - 50y + k = 0 is 7, then what is the value of k ?
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Given quadratic equation is
7y2 - 50y + k = 0
If one root is 7, then it will satisfy the equation i.e putting y = 7 in equation
7 x (7)2 - 50 x 7 + k = 0Correct Option: A
Given quadratic equation is
7y2 - 50y + k = 0
If one root is 7, then it will satisfy the equation i.e putting y = 7 in equation
7 x (7)2 - 50 x 7 + k = 0
⇒ 7 x 49 - 350 + k = 0
⇒ 343 - 350 + k = 0
∴ k = 7
- For what value of k, the equation x2 + 2(k - 4) x + 2k = 0 has equal roots ?
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Given equation is
x2 + 2(k - 4)x + 2k = 0
On comparing with ax2 + bx + c = 0
Here, a = 1, b = 2(k -4), c = 2k
Since, the root are equal, we have D = 0.
b2 - 4ac = 0Correct Option: B
Given equation is
x2 + 2(k - 4)x + 2k = 0
On comparing with ax2 + bx + c = 0
Here, a = 1, b = 2(k -4), c = 2k
Since, the root are equal, we have D = 0.
b2 - 4ac = 0
∴ 4(k - 4)2 - 8k = 0
4(k2 + 16 - 8k) - 8k = 0
⇒ 4k2 + 64 - 32k - 8k = 0
⇒ 4k2 - 40k + 64 = 0
⇒ k2 - 10k + 16 = 0
⇒ k2 - 8k - 2k + 16 = 0
⇒ k(k - 8) -2 (k - 8) = 0
⇒ (k - 8) (k - 2) = 0
Hence, the value of k 8 or 2.
- The quadrictic equation whose roots are 3 and -1, is
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Given that, the roots of the quadrictic equation are 3 and -1.
Let α = 3 and β = -1
Sum of roots = α + β = 3 - 1 = 2
Products of roots = α . β = (3) (-1) = -3
∴ Required quadric equation is
x2 - (α + β)x + α β = 0Correct Option: B
Given that, the roots of the quadrictic equation are 3 and -1.
Let α = 3 and β = -1
Sum of roots = α + β = 3 - 1 = 2
Products of roots = α . β = (3) (-1) = -3
∴ Required quadric equation is
x2 - (α + β)x + α β = 0
⇒ x2 - (2)x + (-3) = 0
⇒ x2 - 2x - 3 = 0
- If α and β are the roots of the equation 4x2 - 19x + 12 = 0, find the equation having the roots 1/α and 1/β
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Given equation is 4x2 - 19x + 12 = 0
Let given equation having the roots 1/α and 1/β,
Then required equuation is
12x2 - 19x + 4 = 0Correct Option: B
Given equation is 4x2 - 19x + 12 = 0
Let given equation having the roots 1/α and 1/β,
Then required equuation is
12x2 - 19x + 4 = 0