quadratic equations Model Questions & Answers, Practice Test for ibps so prelims 2023
ibps so prelims 2023 SYLLABUS WISE SUBJECTS MCQs
Number System
Simplification
Ratio & Proportion
Profit & Loss
Time & Work
Simple Interest & Compound Interest
Mensuration: Area & Volumes
Quadratic Equations
If one the roots of the equation $px^2$ + qx + r = 0 is three times the other, then which one of the following relations is correct ?
Answer: (d)
Given equation $px^2$ + qx + r = 0.
Let one root of the equation be α.
So, other roots = 3 α
∴ Sum of roots = α + 3 α = - $q/p$
⇒ 4 α = ${-q}/p ⇒ α = {-q}/{4p}$.......(i)
Product of roots = (α).(3 α) = $r/p$
Put the value of α
⇒ $3 α^2 = r/p ⇒ 3({-q}/{4p})^2 = r/p$
⇒ ${3q^2}/{16p^2} = r/p ⇒ 3q^2 p = 16p^2 r$
⇒ $3q^2$ = 16pr
The expression $2x^3 + x^2$ - 2x - 1 is divisible by
Answer: (b)
f(x) = $2x^3 + x^2$ - 2x - 1
= $x^2(2x + 1) - 1(2x + 1) = (2x + 1) (x^2 - 1)$
= (2x + 1) (x + 1) (x - 1)
So given expression is divisible by 2x + 1.
If $x^2 = 6 + √{6 + √{6 + √{6 + ..... ∞}}},$ then what is one of the values of x equal to ?
Answer: (c)
$x^2 = 6 + √{6 + √{6 + √{6 + .....∞}}}$
$x^2 = 6 + √{x^2}$
⇒ $x^2$ = 6 + x
⇒ $x^2$ - x - 6 = 0
⇒ $x^2$ + 2x - 3x - 6 = 0
⇒ x(x + 2) - 3(x + 2) = 0
⇒ (x - 3) (x + 2) = 0
∴ x = 3
Alternate Method:
Given,
$x^2 = 6 + √{6 + √{6 + √{6 + ..... ∞}}}$
x = $√{6 + √{6 + √{6 + √{6 + ......∞}}}}$
Here, factor of 6 = 2 and 3
Sign in expression is positive.
Sothat x = 3.
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. $x^2$ + 7x + 12 = 0
II. $y^2$ + 6y + 8 = 0
Answer: (e)
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.
Directions :
In the following 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. $3/{√x} + 4/{√x} = √x$
II. $y^2 = {(7)^{5/2}}/{√y}$ = 0
Answer: (e)
I. ${3}/{√x} + {4}/{√x} = √x⇒{7}/{√x} = √x$
⇒7 = $√{x^2}$ = x ∴ x = 7
II. $y^2 - {(7)^{5/2}}/{√y} = 0⇒ y^2 = {(7)^{5/2}}/{√y}$
⇒$7^2 × √y = (7)^{5/2}⇒(y)^{5/2} = (7)^{5/2}$⇒y = 7
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