## Demo

Q1. If -4 is a solution for x in the equation x^2+kx+8=0, what is k?
Ans. If -4 is a solution for x in the equation.
Then, (-4)^2+k(-4)+8=0
16-4k+8=0
-4k=-24
k=6

Q2. Given that x^2-13x=30, what is x?
Ans. x^2-13x=30
x^2-13x-30=0
x=((-b)±√(b^2-4ac))/2a
=(13±√(169-4(-30) ))/2
=(13±√289)/2
=(13±17)/2
=-4/2 or 30/2
=-2 or 15

Q3. If the area of a certain square (expressed in square meters) is added to its perimeter (expressed in meters), the sum is 77. What is the length of a side of the square?
Ans. Let side of square be x.
Area if square x^2
Perimeter of square = 4x
x^2+4x=77
x^2+4x-77=0
α,β=(-4±√(16-4(-77) ))/2
α=(-4+√(16+4(77)))/2,β=(-4-√(16+4(77) ))/2
α=(-4+18)/2,β=(-4-18)/2
α=7,β=-11
Since the ride of a square must be positive, negative value (-11) has been discorded.
Side of the square = 7

Q4. If x^2-6x-27=0 & y^2-6y-40=0, what is the maximum value of x+y?
Ans. x^2-6x-27=0
α,β=(-b±√(b^2-4ac))/2a
α=(-b+√(b^2-4ac))/2a,β=(-b-√(b^2-4ac))/2a
α=(6+√(36-4(-27) ))/2,β=(6-√(36-4(-27) ))/2
=(6+√144)/2,=(6-√144)/2
α=(6+12)/2=9,β=(6-12)/2=-3
x may be 9 or -3.
y^2-6y-40=0
α,β=(-b±√(b^2-4ac))/2a
α=(-b+√(b^2-4ac))/2a,β=(-b-√(b^2-4ac))/2a
α=(6+√(36-4(-40)))/2,β=(6-√(36-4(-40)))/2
α=(6+14)/2,β=(6-14)/2
α=10,β=-4
y may be 10 or -4.
For maximum value of (x+y),
x should be maximum & y should be minimum.
Maximum value of (x+y) = 9+10 = 19

Q. xy>0
Quantity A: (x+y)^2
Quantity B: (x-y)^2
Ans. Quantity A: (x+y)^2
x^2+y^2+2xy
Quantity B: (x-y)^2
x^2+y^2-2xy
Now subtract x^2+y^2 from both columns.
Then, Quantity A: 2xy
Quantity B: -2xy
xy>0
xy is positive, so the value in quantity A will be positive, regardless of the values in Quantity B will always be negative, regardless of the values of x & y.
Quantity A is larger.

Q. Find the value of √(6+√(6+√(6………)) )
a) 4
b) 3
c) 3.5
d) 2.5
Ans. Let y=√(6+√(6+√(6………)) )
y=√(6+y) (squaring on both sides)
y^2=6+y
y^2-y-6=0
α=(1+√(+1-4(-6) ))/2,β=(1+√(+1-4(-6) ))/2
α=(1+5)/2,β=(1-5)/2
α=3,β=-2
y=3 (Since √(6+√(6+√(6………)) ) cannot be negative.

Q. If the roots of the equation x^2-bx+c=0, differ by 2, then which of the following is true?
a) c^2=4(c+1)
b) b^2=4(c+1)
c) c^2=b+4
d) b^2=4(c+2)
Ans. x^2-bx+c=0
Let α & β be the roots of equation.
α+β=b,dβ=c
α-β=2
(α+β)^2=α^2+2αβ+β^2 (Subtract (+4αβ) from both sides)
(α+β)^2-4αβ=α^2+2αβ+β^2-4αβ
(α+β)^2-4αβ=α^2+2αβ+β^2=(α-β)^2
b^2-4c=(2)^2=4
b^2=4c+4=4(c+1)
〖∴b〗^2=4(c+1)

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