Quadratic Equation
- If one root of the quadratic equation 2x2 + Px + 4 = 0 is 2, find the second root and value of P.
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Here ,The given equation is 2x2 + Px + 4 = 0 .......( ⅰ )
⇒ P(x) = 0 , where P(x) = 2 2 + Px + 4 = 0
If 2 is a root of P(x) = 0, then P(2) = 0
Putting x = 2 in equation ( ⅰ ) , we get
⇒ 2(2)2 + P(2) + 4 = 0
⇒ 8 + 2P + 4 = 0
⇒ 2P = - 12⇒ P = -12 = - 6 2 Correct Option: A
Here ,The given equation is 2x2 + Px + 4 = 0 .......( ⅰ )
⇒ P(x) = 0 , where P(x) = 2 2 + Px + 4 = 0
If 2 is a root of P(x) = 0, then P(2) = 0
Putting x = 2 in equation ( ⅰ ) , we get
⇒ 2(2)2 + P(2) + 4 = 0
⇒ 8 + 2P + 4 = 0
⇒ 2P = - 12
Hence the given equation is⇒ P = -12 = - 6 2
2x2 - 6x + 4 = 0 { ∴ Putting the value of P in eq. ( ⅰ ) }
⇒ 2x2 - 2x - 4x + 4 = 0
⇒ 2x ( x - 1 ) - 4( x - 1 ) = 0
⇒ 2( x - 2 ) ( x - 1 ) = 0
⇒ x = 1 or x = 2
Hence second root is 1.
- The roots of the equation 3a2x2 - abx - 2b2 = 0 are
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According to question ,
The given quadratic equation is
3a2x 2 - abx - 2b 2 = 0 ⇒ 3a 2 x 2 - 3abx + 2abx - 2b 2 = 0
⇒ 3ax ( ax - b ) + 2b ( ax - b ) = 0
⇒ ( ax - b ) ( 3ax + 2b ) = 0
⇒ ax = b or 3ax = - 2b⇒ x = b or x = - 2b a 3a Correct Option: A
According to question ,
The given quadratic equation is
3a2x2 - abx - 2b 2 = 0 ⇒ 3a 2 x 2 - 3abx + 2abx - 2b 2 = 0
⇒ 3ax ( ax - b ) + 2b ( ax - b ) = 0
⇒ ( ax - b ) ( 3ax + 2b ) = 0
⇒ ax = b or 3ax = - 2b⇒ x = b or x = - 2b a 3a
Hence , option A will be required answer .
- The roots of the equation ax 2 + ( 4a 2 - 3b )x - 12ab = 0 are
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Let k1 and k2 be the roots of given quadratic equation .
The given quadratic equation is
ax2 + ( 4a2 - 3b )x - 12ab = 0
⇒ ax2 + 4a2x - 3bx - 12ab = 0
⇒ ax ( x + 4a ) - 3b ( x + 4a ) = 0
⇒ ( x + 4a ) ( ax - 3b ) = 0
⇒ ax - 3b = 0 or x + 4a = 0⇒ x = 3b or x = - 4a a Correct Option: C
Let k1 and k2 be the roots of given quadratic equation .
The given quadratic equation is
ax2 + ( 4a2 - 3b )x - 12ab = 0
⇒ ax2 + 4a2x - 3bx - 12ab = 0
⇒ ax ( x + 4a ) - 3b ( x + 4a ) = 0
⇒ ( x + 4a ) ( ax - 3b ) = 0
⇒ ax - 3b = 0 or x + 4a = 0⇒ x = 3b or x = - 4a a
Thus, the two roots of the given quadratic equation are- 4a and 3b a
- If α, β are the roots of the equation x2 + kx + 12 = 0 such that α − β = 1, the value of k is
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As per the given above question , we know that
Let, α, β be the roots of the equation x2 + kx + 12 = 0
∴ α + β = − k and αβ = 12
Now ( α - β )2 = ( α + β )2 - 4αβ
( 1 )2 = k2 - 4(12)Correct Option: D
As per the given above question , we know that
Let, α, β be the roots of the equation x2 + kx + 12 = 0
∴ α + β = −k and αβ = 12
Now ( α - β )2 = ( α + β )2 - 4αβ
( 1 )2 = k2 - 4(12)
⇒ k2 = 49 ⇒ k = ± 7.
- The value of k for which the roots α, β of the equation : x2 − 6x + k = 0 satisfy the relation 3α + 2β = 20, is
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According to question , we have
The roots α, β of the equation : x2 − 6x + k = 0
On comparing with ax2 + bx + c = 0 , we get a = 1 , b = - 6 , c = k
Now , α + β = ( -b/a ) = 6Correct Option: A
According to question , we have
The roots α, β of the equation : x2 − 6x + k = 0
On comparing with ax2 + bx + c = 0 , we get a = 1 , b = - 6 , c = k
Now , α + β = ( - b/a ) = 6
α + β = 6 and 3α + 2β = 20
On solving ,
⇒ α = 4, β = 2
Product of the roots = k
So, k = αβ = 4 × 2 = 8.
Hence , required answer will be option A .
