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
The difference in the roots of the equation $2x^2$ - 11x + 5 = 0 is
Answer: (d)
Let α and β be the roots of this quadratic equation
$2x^2$ - 11x + 5 = 0
∴ α + β = - ${(- 11)}/2 = {11}/2$.......(i)
and α . β = $5/2$
We know that
$(α - β)^2 = (α + β)^2 - 4 α β$
= $({11}/2)^4 -4(5/2) = {121}/4 - {40}/4 = {81}/4 = (9/2)^2$
∴ Difference of roots = (α - β) = $9/2$ = 4.5.
If (b- 6) is one of the quadratic equation $x^2$ - 6x + b = 0, where b is an integer, then what is the maximum value of $b^2$ ?
Answer: (c)
(b - 6) is a root of
$x^2$ - 6x + b = 0
⇒ $(b - 6)^2 - 6(b - 6) + b = 0$
$b^2$ + 36 - 12b - 6b + 36 + b = 0
$b^2$ - 17b + 72 = 0
$b^2$ - 9b - 8b + 72 = 0
b(b - 9) - 8(b - 9) = 0
(b - 8) (b - 9) = 0
⇒ b = 9 or b = 8
Maximum value of $b^2 = 9^2$ = 81
If α and β are the roots of the equation $x^2$ - 6x + 6 = 0, what is $α^3 + β^3 + α^2 + β^2$ + α + β equal to ?
Answer: (b)
α + β = 6 and α β = 6
∴ $(α + β)^2 = 6^2$
⇒ $α^2 + β^2$ + 2 α β = 36
⇒ $α^2 + β^2$ = 36 - 2 (6) = 24
∴ $(α^3 + β^3) + (α^2 + β^2) + (α + β)$
= $(α + β) (α^2 + β^2 - α β) + (α^2 + β^2) + (α + β)$
= 6(24 - 6) + (24) + (6)
= 6(18) + 30 = 108 + 30 = 138
If one root of $px^2$ + qx + r = 0 is double of the other root, then which one of the following is correct ?
Answer: (d)
Given, $px^2$ + qx + r = 0
Let α and β are the roots of equation.
According to the question,
β = 2 α
Product of roots (α β) = $r/p = 2 α^2$
⇒ $α^2 = r/{2p}$
Sum of roots (α + β) = -$q/p$ = 3 α
⇒ α = ${- q}/{3p}$
We put the value pf α in equation (i),
Now,
⇒ $(q/{3p})^2 = r/{2p}$
$2q^2$ = 9pr
The value of y which will satisfy the equations $2x^2$ + 6x + 5y + 1 and 2x + y + 3 = 0 may be found by solving which one of the following equations ?
Answer: (a)
Given that,
$2x^2$ + 6x + 5y + 1 = 0.......(i)
and 2x + y + 3 = 0 ⇒ x = ${-y -3}/2$ .....(ii)
Now, putting the value of x from (ii) in equation (i),
$2({-3 - y}/2)^2 + 6({-3 -y}/2)$ + 5y + 1 = 0
⇒ ${9 + y^2 + 6y}/2 - {(18y + 6y)}/2$ + 5y + 1 = 0
⇒ $y^2$ + 10y - 7 = 0
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