quadratic equations Model Questions & Answers, Practice Test for ibps clerk prelims 2023
ibps clerk prelims 2023 SYLLABUS WISE SUBJECTS MCQs
Ratio & Proportion
Percentages
Profit & Loss
Time & Work
Time & Distance
Simple Interest & Compound Interest
Mensuration: Area & Volumes
Algebraic Expressions
Trigonometric Ratios & Identity
Linear Equations
Quadratic Equations
Logarithm
If α and β are the roots of the equation $ax^2$ + bx + c = 0, then the value of $1/{a α + b} + 1/{a β + b}$ is
Answer: (b)
Equation, $ax^2$ + bx + c = 0 have root α and β then a $α^2$ + b α + c = 0 ⇒ a $α^2$ + b α = -c
⇒ (a α + b) = ${-c}/{α}$
$a β^2 + b β + c = 0 ⇒ a β^2 + b β = -c ⇒ (a β + b) = {- c}/{β}$
Now, $1/{(a α + b)} + 1/{(a β + b)} = - {α}/c - {β}/c$
= - ${(α + β)}/c = - {(- b/a)}/c = b/{ac}$
If α and β are the roots of the quadratic equation $x^2$ + kx - 15 = 0 such that α - β = 8, then what is the positive value of k ?
Answer: (d)
As α and β are roots of the equation $x^2 + kx - 15$ = 0
Then, sum of roots (α + β) = -k,
Product of roots α β = - 15 and (α - β) = 8 (given)
$(α - β)^2 + 4 α β = (α + β)^2$
64 + (4 × - 15) = $k^2$
⇒ $k^2$ = 4 ⇒ k = 2
Directions :
In each of these questions, two equations numbered I and II are given. You have to solve both the equations and –
- if x < y
- if x ≤. y
- if x > y
- if x ≥ y
- if x = y or the relationship cannot be established.
I. $x^2$ + 12x + 32 = 0
II. $y^2$ + 17y + 72 = 0 11.
Answer: (c)
Note: Let the quardatic equation be $ax^2$ + b + c = 0.
To find roots of this equation quickly, we find two factors of 'b' such that their sum is equal to b and their product is equal to the product of the coefficient of x2 and the constant term 'c'.
Let two such factors be α and β.
The α + β = b and α β = ca
In the second step, we divide these factors by the coefficient of $x^2$ ,
ie be 'a'.
In the next step, we change the signs of the outcome. These are the
roots of the equation.
The sum of the roots of the equation $1/{x + a} + 1/{x + b} + = 1/c$ is zero. What is the product of the roots of the equation?
Answer: (a)
Given, $1/{x + a} + 1/{x + b} = 1/c$
⇒ ${(x + b) + (x + a)}/{(x + a) (x + b)} = 1/c$
⇒ 2cx + (a + b)c = $x^2$ + (a + b)x + ab
⇒ $x^2$ + (a + b - 2c)x + ab - ac - bc = 0
Let the roots of above equation be α and β.
Given, α + β = 0
⇒ -a (a + b - 2c) = 0
⇒ a + b = 2c ....(i)
Now, α β = ab - ac - bc = ab - (a + b)c
= ab - (a + b) ${(a + b)}/2$ [from equation (i)]
= ${2ab - (a^2 + b^2 + 2ab)}/{2} = -{(a^2 + b^2)}/2$
If α and β are the roots of the equation $(x^2 - 3x + 2 = 0),$ then which equation has the roots (α + 1) and (β + 1) ?
Answer: (c)
Since, α and β are the roots of the equation
$x^2$ - 3x + 2 = 0
∴ α + β = 3 and α β = 2 ....(i)
Now, α + 1 + β + 1 = α + β + 2
= 3 + 2 = 5
and (α + 1) (β + 1) = α β + α + β + 1
2 + 3 + 1 = 6
Required equation is
$x^2$ - (α + 1 + β + 1) x + (α + 1) (β + 1) = 0
⇒ $x^2$ - 5x + 6 = 0
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