Quadratic Equation
- x = 3a solution of the equation 3x2 + (k − 1)x + 9 = 0 if k has value
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The given quadratic equation 3x2 + (k − 1)x + 9 = 0
Putting x = 3 in above given eq. , we get
27 + 3(k − 1) + 9 = 0
⇒ 27 + 3k - 3 + 9 = 0
⇒ 3k = −33 ⇒ k = −11.Correct Option: D
The given quadratic equation 3x2 + (k − 1)x + 9 = 0
Putting x = 3 in above given eq. , we get
27 + 3(k − 1) + 9 = 0
⇒ 27 + 3k - 3 + 9 = 0
⇒ 3k = −33 ⇒ k = −11.
Hence , the value of k is −11.
- The expression x2 − x + 1 has :
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According to question , we have
Comparing x2 - x + 1 with ax2 + bx + c , we get
a = 1, b = −1, c = 1
Here D = b2 - 4ac = ( - 1 )2 - 4 (1)(1)
= 1 - 4 = - 3Correct Option: C
According to question , we have
Comparing x2 - x + 1 with ax2 + bx + c , we get
a = 1, b = −1, c = 1
Here D = b2 - 4ac = ( - 1 )2 - 4 (1)(1)
= 1 - 4 = - 3
Since D < 0, so the given expression has no proper linear factor.
- If the equations x2 + 2x - 3 = 0 and x2 + 3x - k = 0 have a common root, then the non - zero value of k is :
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As per the given above question , we can say that
Let, α be a common root of the given equations.
Putting the value of x = α , α2 + 2α - 3 = 0 and α2 + 3α - k = 0
By cross - product method , we get∴ α2 = α = 1 -2k + 9 -3 + k 3 -2 So, α2 = 9 - 2k and α = k - 3 1 1 Correct Option: D
As per the given above question , we can say that
Let, α be a common root of the given equations.
Putting the value of x = α , α2 + 2α - 3 = 0 and α2 + 3α - k = 0
By cross - product method , we get∴ α2 = α = 1 -2k + 9 -3 + k 3 -2 So, α2 = 9 - 2k and α = k - 3 1 1
So, ( 9 - 2k ) = ( k - 3 )2 ⇒ k2 - 4k = 0
⇒ k( k - 4 ) = 0, so k = 4.
Hence , the non-zero value of k is 4 .
- Find the quadratic equation whose roots are reciprocal of the roots of the equation 3x2 - 20x +17 = 0
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The given quadratic equation is
3x2 - 20x +17 = 0 .......(1)
Compare with ax2 + bx + c = 0, we get
a = 3, b = −20, c = 17
The roots of (1) are given byx = - b ± √b2 - 4ac = 20 ± √400 - 4(3)(17) 2a 2 × 3
Correct Option: A
The given quadratic equation is
3x2 - 20x +17 = 0 .......(1)
Compare with ax2 + bx + c = 0, we get
a = 3, b = −20, c = 17
The roots of (1) are given byx = - b ± √b2 - 4ac = 20 ± √400 - 4(3)(17) 2a 2 × 3 x = 20 ± √196 = 20 + 14 , 20 - 14 6 6 6 x = 34 , 6 = 17 , 1 6 6 3 Hence the roots of (1) are 17 and 1. 3 So we have to form an equation whose are 3 and 1 17 Sum of the roots = 3 + 1 = 20 17 17
Hence, the required equation isProduct of the roots = 3 x 1 = 3 17 17 x2 - 20 x + 3 = 0 17 17
⇒ 17x2 - 20x + 3 = 0.
- If log10 ( x2 - 6x + 45 ) = 2, then the values of x are
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Given that : - log10 ( x2 - 6x + 45 ) = 2
⇔ x2 - 6x + 45 = 102 = 100
⇔ x2 - 6x - 55 = 0Correct Option: D
Given that : - log10 ( x2 - 6x + 45 ) = 2
⇔ x2 - 6x + 45 = 102 = 100
⇔ x2 - 6x - 55 = 0
⇔ ( x - 11 ) ( x + 5 ) = 0
⇔ x = 11 or x = - 5.
Hence , the values of x are 11 and - 5.
