Advanced Maths – Equivalent Expressions
1) All the following are equivalent to 3xy = 4ab except:
A) 4a / 3x = y / b
B) 4a / y = 3x / b
C) 4a / b = 3x / y
D) b / y = 3x / 4a
Solution:
A) 4a / 3x = y / b
(Multiply both sides by 3xb, the least common multiple of the denominators)
4a * 3xb / 3x = y * 3xb / b
4ab = 3yx
B) 4a / y = 3x / b
(Multiply both sides by yb, the least common multiple of the denominators)
4a * yb / y = 3x * yb / b
4ab = 3xy
C) 4a / b = 3x / 4a
(Multiply both sides by 4ab, the least common multiple of the denominators)
4a * 4ab / b = 3x * 4ab / 4a
16a * a = 3xb
D) b / y = 3x / 4a
(Multiply both sides by 4ay, the least common multiple of the denominators)
b * 4ay / y = 3x * 4ay / 4a
4ab = 3xy
Correct ANS:
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2) x(y2) – 3y – 3x2y – (-x(y2) + 3x2y – 3Y)
Which of the following is equivalent to the expression?
A) 2x(y2) – 6x2y
B) 6x(y2) – 2x2y
C) 3x(y2) – 5x2y
D) -2x(y2) + 6x2y
Solution:
(x(y2) – 3y – 3x2y) – (-x(y2) + 3x2y – 3Y)
= x(y2) – 3y – 3x2y + x(y2) – 3x2y + 3y
= 2x(y2) – 6x2y
Correct Answer: A
3) Which of the following expressions is equivalent to (p5 y2 q-1)(p2 y3 q3)?
A) p7 y5 q2
B) p3 y7 q2
C) p2 y7 q5
D) p7 y5 q3
Solution:
(p5 y2 q-1)(p2 y3 q3)
= p(5+2) y(2+3) q(-1+3) (open the parentheses)
= p7 y5 q2
Correct Answer: A
4) Which of the following is equivalent to y6 – y3 – 6?
A) (y3 + 3)
B) (y3 – 3)(y3 + 2)
C) (y2 + 3)(y2 – 2)
D) (y3 + 2)(y3 – 3)
Solution:
Factorize using quadratic equation:
y6 – y3 – 6
Hint: Factors of -6 are 2, 3 and 1, 6
-2*3 = -6 and -3*2 = -6
-2 + 3 = 1 and -3 + 2 = -1
Therefore, -3 and 2 are correct factors because their sum is -1 and their product is -6
We can write y6 – 3y3 + 2y3 – 6 = y3(y3 – 3) + 2y3(y3 – 3)
Simply (y3 – 3)(y3 + 2)
Correct Answer: B
5) If x > 5, which of the following is equivalent to 1 / (1 / (x + 3) + 1 / (x + 5))?
A) ((x + 3)(x + 5)) / (2x + 8)
B) ((x – 3)(x + 5)) / (2x – 8)
C) ((x + 3)(x + 3)) / (2x + 8)
D) ((x – 3)(x – 5)) / (2x + 8)
Solution:
In the above equation’s denominator:
1 / (x + 3) + 1 / (x + 5)
Simply by multiplying by the least common multiple of the denominator, which is (x + 3)(x + 5)
The denominator becomes:
(1 / (x + 3)(x + 3)(x + 5) + 1 / (x + 5)(x + 3)(x + 5)) / ((x + 3)(x + 5))
= (x + 5 + x + 3) / ((x + 3)(x + 5))
= (2x + 8) / ((x + 3)(x + 5))
The given equation becomes:
1 / ((2x + 8) / ((x + 3)(x + 5))) = 1 ÷ (2x + 8) / ((x + 3)(x + 5)) = 1 × ((x + 3)(x + 5)) / (2x + 8)
= ((x + 3)(x + 5)) / (2x + 8)
Correct Answer: A
Advanced Maths – Equivalent Expressions
6) If ∛(x5 ) / ∛(x2) is equivalent to x(a⁄b), what is the value of a/b?
A) 1
B) -1
C) 1/2
D) -1/2
Solution:
The given equation: ∛(x5) / ∛(x2)
= x(5/3) / x(2/3)
= x(5/3 – 2/3)
=x3/3
=x1
Therefore, a/b = 1
Correct Answer: A
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7) If x (1 – r) = 1 and r = ky and x ≠ 1, which of the following is equal to y?
A) (x – 1) / kx
B) (x + 1) / kx
C) kx / (x – 1)
D) kx / (x + 1)
Solution:
x (1 – r) = 1
x – rx = 1
(Subtract 1 from both sides)
x – rx – 1 = 1 – 1
x – rx – 1 = 0
x – rx = 1 (rearrange)
(Add rx to both sides)
x – rx + rx = 1 + rx
x = 1 + rx (rearrange x on one side)
rx = 1 – x
r = (1 – x) / x and it is given that r = ky
By equating the value of r, we get ky = (1 – x) / x
Therefore, y = (1 – x) / (xk)
(divide both side by k)
Correct Answer: A
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Non-linear Functions
1) Area of a triangle A =1/2 × Base ×height ,where base and height of the triangle are in the ratio 5: 3 . and base and height represented by b and h find the ares of the triangle in terms of its height.
A) 6/5 h2
B) 5/6 h2
C) 5/6 b2
D) 3/10 b2
Solution:
Area of triangle = 1/2×base×height =1/2×b×h = 1/2 bh
The ratio of base : height = b:h = 5:3
Therefore 5h = 3b Divide 5 on both side
Divide both sides by 5
h = 3b/5
substitute h =3b/5 in area
So area of the triangle = 1/2×b×3b/5=3/10×b2
Correct Answer: D
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2) The function f(x) defined as f(x) = (x-1)2 +(x+1)2+10
In xy plane, the points (0, 0) and (t, 0) are on the graph. Then find the value of t.
A) 2
B) 3
C) -2
D) -3
Solution:
Type equation here.
Function f(x) = (x-1)2 + (x+1)2 – 10
Substitute (t, 0) in the above equation
(t-1) 2 + (t+1) 2 – 10 = 0
t2 – 2t + 1 + t2 + 2t + 1 – 10 = 0
2t2 + 2 – 10 = 0
2t2 – 8 = 0 (divide by 2 on both sides)
t2 – 4 = 0
(t + 2) (t – 2) = 0 (factorize)
So t – 2 = 0, t + 2 = 0
The value of t = 2, t = -2
Consider the positive value.
Correct Answer: A
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3) The function f(x) as defined as f(x) = 5ex – 15. The graph f(x) is in the xy-plane. Find the y-intercept of the above function.
A) -10
B) 10
C) 15
D) -15
Solution:
To find the y-intercept, substitute x = 0 in the given function.
f(0) = 5e0 – 15 = 5 × 1 – 15 = 5 – 15 = -10
Correct Answer: A
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TRANSLATION OF QUADRATIC FUNCTION.:
Horizontal translation:
A function translated to Right or Left Is called Horizontal Translation:
Vertical Translation:
A function is translated to up or down is called Vertical translation.
Horizontal translation:
A function translated to Right or Left Is called Horizontal Translation:
Vertical Translation:
A function is translated to up or down is called Vertical translation.
Examplers:
!) The function f (x) =x2 +4x +6 graphed in XY plane, where y = f ),
Is translated 2 units right. Which equation defined the translated function g (x).
A) x2 +8
B) x2-8 x + 18
C) x2 – 8 x – 18
solution
Function f ( x) = x2 + 4x +6
It is translated to right by 2 units
Rule: Any function f(x) is translated to right by “ h “
Then the translated equation of f ( x) is f( x-h).
Replace “ X” in f(x ) by “x- 2”
Therefore the translated equation of f ( x) is g( x )
G ( x ) = (x-2)2– 4 (x-2) +6
= x2 – 2×24 – 4x +8 +6 = x2 – 4x +4 -4x + 8 +6
= x2 -4x +4 – 4 x +14 = x2 – 8 x +18
Correct ANS : B
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2) The function f (x) =x2 +4x +6 graphed in XY plane , where y = f (x), Is translated 2 units right. Which equation defined the translated function g (x).
A) x2 +2 x + 3
B) x2-8 x + 18
C) x2 – 2 x – 18
D) x2– 8 x +3
Solution:
Function f ( x) = x2 + 4x +6
It is translated to left by 3 units
Rule: Any function f(x) is translated to right by “h “
Then the translated equation off (x) is f( x+h).
Replace “ X” in x) by “x + 3”
Therefore the translated equation off (x)
G (x) = 〖(x+3) 〗2– 4 (x+3) +6
= x
= x2 + 2 x + 3
Correct ANS: A
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3) function f (x ) = x2 + 3 x – 2 I , The graph of the function f in XY – Plane , Where
Y = f ( x ) , is translated 5 units down produce the graph of y = g (x).
Which equation define the function g(x)
A)x2 +2x – 7
B) x2+3x-7
C) x2 +3x +7
D)x2-3x-7
Rule: Any function is translated “k “units down
solution
The translated equation is f (x) – k
The given function f (x ) = x2+ 3 x – 2
Is translated to 5 units down.
Therefore the translated equation g (x) = f (x) – K = x2 + 3 x – 2 – 5
g ( x ) = x2 + 3 x – 7
Correct ANS: B
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4) function f ( x ) =2 x2 – 6 x – 6, The graph of the function f in XY – Plane, Where
Y = f (x ) , is translated 4 units up produce the graph of y = g (x ).
Which equation define the function g ?
A)2x2 -6x -2
B) x) < sup>2-6x+2
C)2 x2 +3x +7
D)x2-3x-7
Rule : Any function is translated k units up The translated equation is f (x) +k
solution:
The given function f (x ) = 〖2x〗2
– 6 x – 6Is translated to 4 units up
Therefore the translated equation g (x) = f (x) + K =2 x2 -6x – 6 + 4
g ( x ) =2 x2 -6 x – 2
Correct ANS:A
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5) The function f(x) is graphed in XY -plane. The fuction f(x) is translated 5 units up and 3
Units left to produce the graph y=g(x)
which of the following function defined g(x)?
A)9/(x+5)+11
B)9/(x+5)-11
C)9/(x-5)+11
D) (-9)/(x+5) +1
Solution:
solution
The given function f(x) = 9/(x+2)+6
Is translated 5 units up
Therefore the vertical translated graph is =9/(x+2) +6+5 (5 units translated up by 5units)
Now the function f(x) is translated 3 units left.
The new function g(x) =9/(x+2+3) +11
(3 units translated left)
Therefore g(x)= 9/(x+5)+11
Correct ANS: A
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Projectiles;
Projectile of an object is represented by a Quadratic equation
General quadratic equation is represented as ax2+bx +c
Where a, b, and c are constants.
The quadratic equation is open up or down;
In The equation ax2
+ bx + cThe value of a is called the co-efficient of x2
,If it is positive then the quadratic equation open up wards
If it is negative then the quadratic equation is open down words
The quadratic equation has reach minimum or maximum
If the quadratic equation is open up it reaches minimum.
If it is quadratic equation open down words, then it reaches
Maximum
EXAMPLES:
1) the projectile of an object is represented by f(x) =2x2
+25x +62 and it is graphed in XY -plane.which of the following is true for the above projectile
A) Open upwards and it reaches maximum
B) Open downwards and it reaches minimum
C) Open upward and it reaches minimum
Open downward and it reaches maximum
Solution:
The given equation is = 2 x2 +25x +c
Co-efficient of x2 is 2 and it is positive
As per the properties of quadratic equation
The quadratic equation is open upwards, and it reaches minimum.
Correct ANS: C
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2) the projectile of an obj ect is represented by f(x) =-3x2+25x +62 and it is graphed in XY -plane.
which of the following is true for the above projectile
A)Open upwards and it reaches maximum
B)Open downwards and it reaches minimum
C)Open upward and it reaches minimum
C) Open downward and it reaches maximum
Solution:
The given equation is = -3x2 +25x +c
Co-efficient of x2
is -3 and it is negative
As per the properties of Projectiles quadratic equation
The quadratic equation is open downward, and it reaches maximum.
Correct ANS: C
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Sum for Practice:
1) p(x) = -9x2+ 12 x +10 The quadratic function defines p(x)
Which of the following statement is correct.
A) Open upwards and it reaches maximum
B) Open downwards and it reaches minimum
C)Open upward and it reaches minimum
D) Open downward and it reaches maximum
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2) h(x) = 5x^2-2x+6 the given equation represents the function h and it
Is graphed in XY-plane.
Which of the flowing statements are correct.
A) Open upwards and it reaches maximum
B) Open downwards and it reaches minimum
C) Open upward and it reaches minimum
D) Open downward and it reaches maximum
2) h(x) = 5x2-2x+6 the given equation represents the function h and it
Is graphed in XV -plane.
Which of the flowing statements are correct.
A) Open upwards and it reaches maximum
B) Open downwards and it reaches minimum
C)Open upward and it reaches minimum
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Vertex form of a quadratic function
The quadratic function f(x) = ax2+bx+ c can rewrite in the form of,su Vertex
The general vertex form of a quadratic function f(x) can be represented as f(x) = 〖(x-h) 〗2+k
Here (h, k) is called vertex
The vertex is a point at which the quadratic function reaches maximum or minimum.
At this point of the Quadratic function has one solution,
It means the quadratic function has only two equal roots
That is root has only one value.
In the vertex form the value of h and k represents x and y co-ordinates of the quadratic function
The the value of k represents the minimum or minimum value the
Quadratic function reaches
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Examples;
The quadratic function f(x) = 〖(x-5) 〗2+6 is graphed in XY-plane.
Can you write it open upwards or open down wards and it reaches maximum or minimum
Option;
A) Open upwards and it reaches minimum value is 6
B)open down wards and it reaches maximum value is 6
c) open upwards and it reaches minimum value is 6
D) Open downwards it reaches minimum value is 6
Solution :
The given quadratic function f(x) = 〖(x-5) 〗2+6
is in Vertex form.
(5,6) is the co-ordinates of the vertex.
Here the y co-ordinate represents the the minimum or Maximum
value of the function f(x) reaches
But her the quadratic equation open upwards, so it reaches Minimum
Here x co-ordinate in the above Vertex form of quadratic function.
The quadratic function open upwards.
So f(x) reaches minimum, and the value k represent the minimum
Minimum height f(x) reaches, and it is 6
The correct answer is the quadratic function is open upwards
and it Reaches minimum vale is 6
Correct answer: A
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Practice :
The function f(x) = -3〖(x-2)〗2+8 is defined in XY- plane.
What is the maximum or minimum value of f(x).
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Roots of a Quadratic function
Any quadratic has two roots.
But in projectile when the quadratic function reaches its maximum or minimum the function has one root.
That point is called the vertex of the quadric function
In genera for the quadratic function f(x) = ax2+bx+c
The roots are x = (-b±√(b2-4ac))/2a
At the Vertex the quadratic function f(x) has one roots (Equal Roots)
Therefore in above x value √(b2-4ac) =0
Now x = (-b)/2a, it means the quadratic function reaches Its maximum or minimum value a =(-b)/2b
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Examples:
The function x) = 3x2+15x – 25 for what value of x , f(x) reaches its minimum?
A) 2.5
B) 5.2
C) 1.5
D)5.1
Solution:
It is given the function f(x) reaches minimum.
So the function has one root.
By formula explain above the root
Therefore x = (-b)/2a = (-15)/ (2×3) = (-15)/6 = (-5)/2= -2.5
Correct ANS: A
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Practice :
f(x) = -5x2 + 8x + 9 the given equation define the function f.
for what value of x the function reaches its maximum and what is the
function reaches its maximum
Correct ANS: (-4/5) , maximum height =61/5
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2)The function f(x) defined by f(x)=(x+3) (x-5).
For what vale of x it reaches minimum and what is the minimum value of f(x).
A) X co-ordinate of vertex is X=1 and minimum value is 14
B) X co-ordinate of vertex is x=1 and minimum value is -14
C)X co-ordinate of vertex is x=- 1 and minimum value is 14
D) X co-ordinate of vertex is x= – 1 and minimum value is 14
Solution:
The given function f(x) = (x+3) (x-5) =x(x-5) +3(x-5)
= x2-5x+3x-15 = x2-2x-15
Since coefficient of x2 term is positive
And graphed in XY -plane
So it is open Upward and it reaches minimum, (quadratic equation rules)
Function reaches it’s minimum value at its vertex
For the equation ax^2+bx+c ,where a, b and c are constants
X -coordinate of the vertex is (_b)/2a, here a= 1 and b= -2
x-Coordinate of the vertex is –((-2)/2) = 1
So the function f(x) reaches its minimum value when the value of x =1
Now the minimum value of f(x) is
The value of f(x) when x =1 is f(1) = 1 – 2×1+15 =1-2+15 = -1 + 15 = 14
The function reaches it’s minimum value at x=1 and it’s minimum value is 14
Correct ANS = A
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Practice sum
The function f(x) = ( -2x+1) (x+1) is graphed in xy- plane
When it reaches it’s minimum or maximum height.
Ans: it reaches maximum
When x= 1/(4 ) and maximum height is =1/4
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NOTE: Projectile of an object at given height above the ground. This can be represented by f(x) = a(〖x-h)〗2+k K represents the height above ground and x represent the value of f(x) after the time x and h represent the function reaches minimum or maximum.1)An object is launched off from an elevated surface above ground of 5 units above the ground surface and
it reaches its maximum517 units above the ground in 4 units of time after being launched .With reference to the above given data what is the
height the object reached after 6 units of time after it launched .
A) 389
B) 380
C) 489
D) 269
Solution:
As per the above rule the given statement is represented by
The function f(x) =a (〖x-h)〗2+k where h and k are constants
Such that the function reaches its maximum value k when x is equal to h units
Here the vale of k=517 units above the ground after 4 units of time after it is launched form the elevated surface
so h=4 unit of time and it represents a initial height of 5 units
Therefore f(x) = a(〖x-4)〗2+517 apply the initial height
The x value at its initial height = 0
So f(x) =a(〖0-4)〗2+517 =a×16+517=5
16a = 5 – 517 =- 512
So a = – 512/16= – 32
So the above model can be represented as
f(x) = – 32 (〖x-4)〗2+517
now to obtain the height of the object reached in 6 units of time
Substitute x = 6 in the above equation.
So f(x)= -32 (〖6-4)〗2+517=-32 ×2^2+517=-32×4+517
= – 128 + 517 =389
Correct ANS: A
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Practice Sum:
An object is launched off an elevated surface at height of 2 feet above the ground .and it reaches
a maximum height of 20 feet above the ground in 3 seconds . Based on this above model
what is the height it reaches after 5 seconds.
ANS:16 feet
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Exponential function
General exponential function is of the form f(x) =a(〖b)〗x or a(〖b)〗(x/k)
Examples : exponential series
Where a , b and k are constant numbers and greater than zero.
Second type of exponential form is f(x) = a( 1+〖r/100)〗n
Examples of this type is population , production of goods
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Examples:
The first term of a sequence is 5 and each of the following term is twice the preceding term
of the sequence What is the n th term of the sequence,
Option :
A ) 5(2x)
B)2(5x)
c) 2(5(x-1))
D)5(2(x-1) )
Solution:
The first term of the sequence is 5
Second term on wards increased by 2
That is it is represented 2x
The n th term is equal to 5(2(x-1))
Correct ANS: D
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———————————————————2)the exponential growth of a sequence is given in the table.
x 0 1 2 3
y 1 3 9 27
Write the 10 th term of the sequence.
Solution:
The first term of the sequence is 1
Second onwards is increased by 3.
General term of the sequence is = 3(x-1)
So 9 th terms of the sequence is = 3(9-1)=38
Correct ANS: 38
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3)In a function p , the value of p is increased by 15% for every
Every increase in the value of x by 1 . if p(0) = 5.
Which equation defined p?
A)P(x)=5(1.15) x
B)P(x)=(1.15)x
C)P(x)=2(〖2.15)〗x
D)P(x)=-5(1.15) x
Solution:
P(x) is increased by a fixed percentage 15% for every increase in the value of x by 1
So p(x) is increasing by exponential function
The function is written in the form p(x) = a(〖1+r/100)〗x
The a is the value when p(x) =0
That is here the value of a =5
The value of r is the percentage of increase
Here value r is 15 % and r=15
Substitute the value of a and r in the function p(x)
Therefore p(x) = 5 (〖1+15/100)〗x =5(〖1.15)〗x
Correct ANS: A
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4) The growth of grass beetles in particular garden is given as below
There were 10,000 at the beginning of the observation in a squire kilometer area.
The number of beetles per squire kilometer is increased three time for every 5 hours.
How many grass beetles are there in the square kilometer area after 20 hours.
Option :
A) 270000
B) 810000
C) 910000
D) 2430000
Solution:
Let q represent the number of grass beetles per square kilometer
X hours after the initial observation
Since number of grass beetles per square kilometer is increased 3 time for every 5 hours
The above relationship can be represented by p(x) =a(〖b)〗(x/k)
Where a = number of grass beetle at the beginning of observation.
The number of grass beetles increased by b of every hours
Therefore b= 3 and k= 5
So q(x)= 10000 (〖3)〗(x/5)
Number of grass beetles after 20 hours
Is obtained by substituting m x = 20 in p(x)=10000(〖3)〗(20/5)
That is p(x)=10000× (3)4=810000 grass beetles.
correct ANS: B
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Practice:
1)The exponential function f(x)= 19(〖a)〗x
Where a is positive constant
If f(3) =2,375 what is the value of f(4) ?
ANS: f(4) =11,875
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2)An exponential function q(x) is defined as follow.
The q(0)=43 and for every increase in x by 1
The value of f(x) is decreased by 40% and find the value of f(x) for x=3
ANS:9.288
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1)The exponentials function Is f(x) = a(1± b/100 )x when increase or decrease is given Function is given in percentage
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2)The exponential increase of any function
is f(x) = a(b)x when the increase b for every time unit
The exponential increase of any function
Is f(x) =a(b)(x/k) where b units increase in k units of time
Example increased doubled in every 3 hours
Then f(x) =a (2)(x/3)
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1)A quadratic function y = f(x)= (x+a)(x-4) is graphed in XY – plane as above in the diagram.
Where a is a constant. Find the value of the constant.
Option:
A) 1 B) -1 C) 4 D) -4
Solution:
From the diagram the graph of the function y=f(x)= (x+a)(x-4)
Intersects the X-axis at (-1,0)
Now substitute x= -1 and y= 0 in the function y= (x+a) (x-4) =(-1 +a)(-1-4) =0
Therefore (-1+a)(-5)=0
5 – 5a =0
-5a = -5
That is a =(-5)/(-5) =1 ( divide both side by -5)
Correct ANS : A
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Graph of y = (3/5)x - 2
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Nonlinear equation in one variable and system of equations in two variable.
1)Y= p-qx2 In this equation p and q are constants
Y=- r is astright line .where r is constant
Find out how many times the line and parabola intersect.
Option:
A)0
B)1
C)2
D)3
Solution:
Consider the equation y= p- qx2.
The co-officiant of x2 is negative
.
So it open down wards.
Consider the equation y= -r
Here the value of r is negative .
And it is a constant
So y= -r is parallel to x- axis and r is negative
There fore it is below the X-axis ( the negative side , it is in third and fourth quadrant )
Y= - r and y = p – qx intersect in two points,
Correct ANS:C
1)Y= p-qx2 In this equation p and q are constants
Y=- r is astright line .where r is constant
Find out how many times the line and parabola intersect.
Option:
A)0
B)1
C)2
D)3Solution:
Consider the equation y= p- qx2.
The co-officiant of x2 is negative
.So it open down wards.
Consider the equation y= -r
Here the value of r is negative .
And it is a constant
So y= -r is parallel to x- axis and r is negative
There fore it is below the X-axis ( the negative side , it is in third and fourth quadrant )
Y= - r and y = p – qx intersect in two points,
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2) The graph y= 2 x^2-9x intersect the line y=3x at (0,0) and (p,q).
What is the value of p and q?
Option:A)(18 ,6 ) B)(6,18)C )(18,18) D)(6,6)
Solution:
Consider equation y= 2x^2 -9x and y=3x.
Solve this equation
That is substitute the value of y= 3x in y = 2x2- 9.
2x2-9x=3x add – 3x on both side
2x2-9x-3x=3x-3x=0
2x2-12x=0
2x(x-6)=0
If 2x = 0 then the value of x=0
If x-6 =0 then x=6
(-b±√(b^2-4ac))/2a
So y =3×6=18
The value of p=6 and q=18
Correct ANS: B
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3) the line y=2l where l is constant intersect the parabola y= 3 x2-7x at exactly one point.
What is the value of l.?
Solution:
Y= 3x2-7x is also a quadratic equation.
The value of y=2l
3x2-7x=2l, add -2l on both side
Then 3x2-7-2l=2l-2l=0
3x2-7x-2l=0
It is also given that it has only one solution
In the general solution for quadratic equation
X=(-b±√(b2-4ac))/2a
If it has one solution then √(b2-4ac)/2a=0
Her the values of a ,b and c are from the equation y= 3x2-7x
A=3,b=-7 and c=2l
〖(-7)〗2-4×3×2l=0
9-24l=0
24l=9, ( divide both side by 24)
24l/24=9/24
l=3/8
Correct ANS:3/5
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4) find the value of x-7 in the following equation
x2-14x+49=1/9
Option:
A)1/3
B)3
C)1/9
D)1/9
Write quadratic equation in perfect squiring method.
(x-14/2)249+49=1/9
(x-7)2
=1/9√〖(x-7)〗2=√1/9=1/3
x-7=1/3
Correct ANS:A
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5) x2-5x-7 =0 solution of this equation can be written in the form 5/2±√k ,
What is the value of K
Solution:
x2-5x-7=0 is a quadratic equation
By the method of perfect squiring method
(x-5/2)2,/sup>-7-(5/2)2=0
(x-5/2)2-7-25/4=0 add + 7 and +25/4 on both side (x-5/2)2-7-25/4=7+25/4=7×4+25/4 (simplify)(x-5/2)2=28+25/4=53/4 taking the squire root on both side Then we will get √〖(x-5/2)〗2=√53/4 x-5/2=±√53/4=Type equation here. x=5/2±√53/4 Compare 5/2 ±√(k/4) with 5/2 ±√(53/4)
Therefore the value of k=53/4
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6)for the equation (x-P) (x-17) =x-17 solution are given as p ,P+1, 17 the value of p is a constant
which of the above are correct .
A)p and p+1
B)p+1,17
C)1,p
D)17,p
Solution:
(x-p)(x-17)=x-17 is the given equation
Add - (x-17) on both side of the equation.
Therefore (x-P) (x-17) -(x-17) =(x-17) -(x-17) =0
(x-17) [(x-p)-1] =0
Either (x-17) =0 OR ((x-p--1) =0
If x-17=0
Then x =17If (x-p-1) =0
Then x=p+1