CRE 4 - Two Quadratic Equations | Algebra - Quadratic Equations
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Form another quadratic equation whose roots are 2 more than the roots of the quadratic equation 2x2 + 3x + 1 = 0
- (a)
2x2 + 5x - 3 = 0
- (b)
2x2 + 5x + 3 = 0
- (c)
2x2 – 5x - 3 = 0
- (d)
2x2 – 5x + 3 = 0
Answer: Option D
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Explanation :
To form a quadratic equation whose roots are k more than the roots of ax2 + bx + c = 0, we need to substitute x with (x – k) in the original equation.
∴ The required equation is
2(x - 2)2 + 3(x – 2) + 1 = 0
⇒ 2(x2 – 4x + 4) + 3x – 6 + 1 = 0
⇒ 2x2 – 5x + 3 = 0
Hence, option (d).
Workspace:
Form another quadratic equation whose roots are 1 less than the roots of the quadratic equation 2x2 + 3x + 1 = 0
- (a)
2x2 + 7x + 6 = 0
- (b)
2x2 - 7x - 6 = 0
- (c)
2x2 - 7x + 6 = 0
- (d)
2x2 + 7x - 6 = 0
Answer: Option A
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Explanation :
To form a quadratic equation whose roots are k less than the roots of ax2 + bx + c = 0, we need to substitute x with (x + k) in the original equation.
∴ The require equation is
2(x + 1)2 + 3(x + 1) + 1 = 0
⇒ 2(x2 + 2x + 1) + 3x + 3 + 1 = 0
⇒ 2x2 + 7x + 6 = 0
Hence, option (a).
Workspace:
Form another quadratic equation whose roots are twice the roots of the quadratic equation 2x2 + 3x + 1 = 0
- (a)
x2 - 3x - 2 = 0
- (b)
x2 - 3x + 2 = 0
- (c)
x2 + 3x + 2 = 0
- (d)
x2 + 3x - 2 = 0
Answer: Option C
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Explanation :
To form a quadratic equation whose roots are k times the roots of ax2 + bx + c = 0, we need to substitute x with x/k in the original equation.
∴ The require equation is
2(x/2)2 + 3(x/2) + 1 = 0
⇒ x2/2 + 3x/2 + 1 = 0
⇒ x2 + 3x + 2 = 0
Hence, option (c).
Workspace:
If α and β are the roots of the equation 2x2 + 3x + 1 = 0, then form another quadratic equation whose roots are α2 and β2
- (a)
4x2 – 5x + 1 = 0
- (b)
4x2 + 5x - 1 = 0
- (c)
4x2 – 5x - 1 = 0
- (d)
4x2 + 5x + 1 = 0
Answer: Option A
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Explanation :
α and β are the roots of the equation 2x2 + 3x + 1 = 0
∴ Sum of the roots = α + β = -3/2 …(1)
& product of the roots = αβ = ½ …(2)
Now, a quadratic equation whose roots are α2 and β2 is
x2 – (α2 + β2)x + α2 × β2 = 0
Now, α2 + β2 = (α + β)2 - 2αβ = 9/4 – 1 = 5/4
∴ The required quadratic equation is:
x2 – (α2 + β2)x + (αβ)2 = 0
⇒ x2 – 5x/4 + 1/4 = 0
⇒ 4x2 – 5x + 1 = 0
Hence, option (a).
Workspace:
Form another quadratic equation whose roots are reciprocal of the quadratic equation 3x2 - 11x + 13 = 0.
- (a)
x - 11/x + 3 = 0
- (b)
3x2 - 11x + 13 = 0
- (c)
13x2 - 11x + 3 = 0
- (d)
13x2 + 11x + 3 = 0
Answer: Option C
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Explanation :
To form a quadratic equation whose roots are reciprocal of the roots of ax2 + bx + c = 0, we substitute x with 1/x in this equaiton.
∴ The required equation is
3 - 11 + 13 = 0
⇒ 3 - 11x + 13x2 = 0
⇒ 13x2 - 11x + 3 = 0
Hence, option (c).
Workspace:
Form a quadratic equation whose roots are equal in magnitude but opposite in sign to the roots of 2x2 - 3x - 6 = 0
- (a)
2x2 + 3x - 6 = 0
- (b)
2x2 + 3x + 6 = 0
- (c)
-2x2 + 3x - 6 = 0
- (d)
2x2 - 3x + 6 = 0
- (e)
None of these
Answer: Option A
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Explanation :
To form a quadratic equation whose roots are equal in magnitude but opposite in sign to the given quadratic equations, we need to substitute x with -x in the given quadratic equation.
∴ The new quadratic equation is 2(-x)2 - 3(-x) - 6 = 0
⇒ 2x2 + 3x - 6 = 0
Hence, option (a).
Workspace:
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