Download Grade 10 Quadratic Equations

ID : ww-10-Quadratic-Equations [1]
Grade 10
Quadratic Equations
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Answer t he quest ions
(1)
(2)
Solve quadratic equation
= 0 using f actorization.
T he sum of square of two positive numbers is 80. If square of the larger number is 16 times the
smaller number, f ind the numbers.
(3)
Solve quadratic equation
(x ≠ 0 and x ≠ -1).
Choose correct answer(s) f rom given choice
(4) Which of the f ollowing quadratic equations has two distinct real roots?
(5)
a. -3x2 + 5x - 2 = 0
b. -3x2 + 4x - 2 = 0
c. -3x2 + 6x - 4 = 0
d. 2x2 + 4x + 2 = 0
Which of the f ollowing is a solution to inequality x2 - 7 x < - 10
a. 2 < x < 5
b. 2 ≤ x < 5
c. 2 < x ≤ 5
d. 2 ≤ x ≤ 5
(6) Which of the f ollowing equations has 2 as one of its roots?
a. b2 - 6b + 8 = 0
b. b2 - 2b + 1 = 0
c. b2 - 16 = 0
d. b2 + 5b + 6 = 0
(7) T he sum of the squares of two consecutive natural numbers is 421. Find the numbers.
(8)
a. 14 and 15
b. 13 and 14
c. 16 and 17
d. 15 and 16
T he sum of two numbers is 20. If sum of their reciprocals is 5/24, f ind the numbers.
a. 12 and 8
b. 7 and 13
c. 9 and 11
d. 10 and 10
(9) Age of a f ather is equal to the square of his daughter's age. If one year ago, f ather's age was 7
times the daughter's age. Find their current ages.
a. 64 years and 8 years
b. 25 years and 5 years
c. 36 years and 6 years
d. 16 years and 4 years
(10) Find roots of quadratic equation
a.
c. -
and and 4
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= 0.
b. -
and
d. 3 and Personal use only, commercial use is strictly prohibited
ID : ww-10-Quadratic-Equations [2]
(11)
If
3
is a root of the quadratic equation kb 2 - 14b + 12 = 0, f ind the value of k.
2
a. 6
b. 3
c. 5
d. 4
(12) A natural number, when increased by 5, equals 176 times its reciprocal. Find the number.
a. 13
b. 11
c. 14
d. 9
(13) T he sum of the n consecutive natural numbers starting f rom 2 is 54. Find the value of n.
a. 11
b. 9
c. 10
d. 8
(14) Find the roots of the quadratic equation
= 0.
a.
and
b.
and
c.
and
d.
and
(15) Find values of k f or which the quadratic equation 5x2 – kx + 7k = 0 has equal roots.
a. 0 only
b. 0 and 70
c. 0 and 140
d. 140 only
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ID : ww-10-Quadratic-Equations [3]
Answers
(1)
x = -p/(p + q) or, x = -(p + q)/p
Step 1
LHS =
Step 2
=
Step 3
=
Step 4
Hence x = -p/(p + q) or, x = -(p + q)/p
(2)
8 and 4
Step 1
Let smaller number be x. T heref ore square of larger number = 16x.
Step 2
x2 + 16x = 80
Step 3
x2 + 16x - 80 = 0
Step 4
x2 + 20x - 4x - 80 = 0
Step 5
x (x + 20) - 4 (x + 20) = 0
Step 6
(x + 20) (x - 4) = 0
Step 7
x = 4 or -20. Since numbers are positive, smaller number x = 4.
Step 8
Larger number = √(16 x 4) = 8
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ID : ww-10-Quadratic-Equations [4]
(3)
x = 2 or, x = -3
Step 1
On adding two f ractions on LHS,
Step 2
6 (x2 + x2 + x + 1) = 13 (x2 + x)
Step 3
x2 + x - 6 = 0
Step 4
x2 + 3 x - 2 x - 6 = 0
Step 5
x(x + 3 x) - 2 (x + 3) = 0
Step 6
(x - 2) (x + 3) = 0
Step 7
Hence x = 2, or x = -3
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ID : ww-10-Quadratic-Equations [5]
(4) a. -3x2 + 5x - 2 = 0
Step 1
Let's assume, ax2 + bx + c = 0 is a quadratic equation.
D = b2 - 4ac.
If in a quadratic equation, D < 0, then the quadratic equation has no real roots.
If in a quadratic equation, D > 0, then the quadratic equation has two distinct real roots.
If in a quadratic equation, D = 0, then the quadratic equation one real root.
Let's check all of the quadratic equations f or real roots.
Step 2
-3x2 + 5x - 2 = 0
Here, a = -3, b = 5 and c = -2
Now, D = b2 - 4ac
= (5)2 - 4(-3)(-2)
=1
Since, D > 0, the quadratic equation -3x2 + 5x - 2 = 0 has two distinct real roots.
Step 3
-3x2 + 4x - 2 = 0
Here, a = -3, b = 4 and c = -2
Now, D = b2 - 4ac
= (4)2 - 4(-3)(-2)
= -8
Since, D < 0, the quadratic equation -3x2 + 4x - 2 = 0 has no real roots.
Step 4
-3x2 + 6x - 4 = 0
Here, a = -3, b = 6 and c = -4
Now, D = b2 - 4ac
= (6)2 - 4(-3)(-4)
= -12
Since, D < 0, the quadratic equation -3x2 + 6x - 4 = 0 has no real roots.
Step 5
2x2 + 4x + 2 = 0
Here, a = 2, b = 4 and c = 2
Now, D = b2 - 4ac
= (4)2 - 4(2)(2)
=0
Since, D = 0, the quadratic equation 2x2 + 4x + 2 = 0 has one real root.
Step 6
T hus, the quadratic equation -3x2 + 5x - 2 = 0 has two distinct real roots.
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ID : ww-10-Quadratic-Equations [6]
(5)
a. 2 < x < 5
Step 1
First lets f ind the value of x f or which x2 - 7 x = - 10, or x2 - 7 x + 10 = 0
x2 - 7 x + 10 = 0
(x - 2) (x - 5) = 0
x = 2 or x = 5
Step 2
Now you can notice that inequality is satisf ied f or x > 2 and x < 5.
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ID : ww-10-Quadratic-Equations [7]
(6) a. b2 - 6b + 8 = 0
Step 1
T o check if 2 is a valid root of an equation, we need to replace variable in equation by 2 and
see if equation is satisf ied or not. Lets try this f or all f our equations.
Step 2
b2 - 6b + 8 = 0
On putting b = 2 we get,
L.H.S = (2)2 - 6(2) + 8
=0
Now, L.H.S = R.H.S, theref ore 2 is the root of the equation b2 - 6b + 8 = 0.
Step 3
b2 - 2b + 1 = 0
On putting b = 2 we get,
L.H.S = (2)2 - 2(2) + 1
=1
Now, L.H.S ≠ R.H.S, theref ore 2 is not the root of the equation b2 - 2b + 1 = 0.
Step 4
b2 - 16 = 0
On putting b = 2 we get,
L.H.S = (2)2 - 16
= -12
Now, L.H.S ≠ R.H.S, theref ore 2 is not the root of the equation b2 - 16 = 0.
Step 5
b2 + 5b + 6 = 0
On putting b = 2 we get,
L.H.S = (2)2 + 5(2) + 6
= 20
Now, L.H.S ≠ R.H.S, theref ore 2 is not the root of the equation b2 + 5b + 6 = 0.
Step 6
T hus, the equation b 2 - 6b + 8 = 0 has 2 as one of the root.
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ID : ww-10-Quadratic-Equations [8]
(7) a. 14 and 15
Step 1
Let the numbers be x and x+1
Step 2
x2 + (x+1)2 = 421
Step 3
2x2 + 2x + 1 = 421
Step 4
2x2 + 2x - 420 = 0
Step 5
x2 + x - 210 = 0
Step 6
x2 + 15x - 14x - 210 = 0
Step 7
x(x + 15) - 14(x + 15) = 0
Step 8
(x - 14) (x + 15) = 0
Step 9
x = 14 or -15. Since x cannot be negative, x = 14. Hence numbers are 14 and 15.
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ID : ww-10-Quadratic-Equations [9]
(8)
a. 12 and 8
Step 1
Let the numbers are x and 20-x. T hen
Step 2
On adding two f ractions on LHS
Step 3
480 = 5x (20 - x)
Step 4
5 x2 - 100 x + 480 = 0
Step 5
x2 - 20 x + 96 = 0
Step 6
x2 - 12 x - 8 x + 96 = 0
Step 7
x (x - 12) - 8 (x - 12) = 0
Step 8
(x - 12) (x - 8) = 0
Step 9
x = 12 or 8. Hence numbers are 12 and 8.
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ID : ww-10-Quadratic-Equations [10]
(9) c. 36 years and 6 years
Step 1
Let the age of the son = x, hence age of f ather = x2
Step 2
One year ago,
x2 - 1 = 7(x - 1)
Step 3
x2 - 7x + 6 = 0
Step 4
x2 - 6x - x + 6 = 0
Step 5
x (x - 6) - (x + 6) = 0
Step 6
(x - 1) (x - 6) = 0
Step 7
x = 1 or 6. Age cannot be 1 (that would make age of f ather and daughter to be same)
Step 8
T heref ore age of daughter = x = 6 years. Age of f ather = x2 = 36 years
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ID : ww-10-Quadratic-Equations [11]
(10) a.
and Step 1
On comparing the quadratic equation
= 0, with the standard f orm ax2 +
bx + c = 0, we get:
a = 1,
b=
,
c = -24
Step 2
D = b2 - 4ac
)2 - 4(1)(-24)
=(
= 2 + 96
= 98
Step 3
By using the quadratic f ormula, we get:
-b ∓ √D
x=
2a
=
∓
-
2×1
∴x=
or -
Step 4
T hus, the roots of the quadratic equation
= 0, are
and -
.
(11) d. 4
Step 1
It is given that,
3
is the root of the quadratic equation kb 2 - 14b + 12 = 0. T heref ore
2
equation kb 2 - 14b + 12 = 0, should satisf y if we replace b by
3
2
Step 2
kb2 - 14b + 12 = 0
⇒ k(
3
2
)2 - 14(
3
) + 12 = 0
2
Step 3
On solving above equation f or k, we f ind that k = 4
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ID : ww-10-Quadratic-Equations [12]
(12) b. 11
Step 1
Let's assume that the natural number is x.
T he reciprocal of x = 1/x.
Step 2
It is given that when the natural number x is increased by 5, it equals 176 times its
reciprocal.
We can write this f act as an equation and solve f or x as:
x + 5 = 176(1/x)
⇒ x + 5 = 176/x
⇒ x2 + 5x = 176
⇒ x2 + 5x - 176 = 0
Step 3
Let us now solve the equation x2 + 5x - 176 = 0 by the f actorization method:
x2 + 5x - 176 = 0
⇒ x2 + 16x - 11x - 176 = 0
⇒ x(x + 16) - 11(x + 16) = 0
⇒ (x + 16)(x - 11) = 0
either,
|
or,
(x + 16) = 0
|
(x - 11) = 0
⇒ x = -16
|
⇒ x = 11
x ≠ -16, x is a natural number.
T heref ore, x = 11
Step 4
T hus, the natural number is 11.
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ID : ww-10-Quadratic-Equations [13]
(13) b. 9
Step 1
2 + 3 + 4 + .... n terms = 54
Step 2
n(2 × 2 + n - 1)/2 = 54 [Using AP summation Sn = n/2(2a + n - 1 )]
Step 3
n2 + 3n = 108
Step 4
n2 + 3n - 108 = 0
Step 5
(n - 9) (n + 12) = 0
Step 6
n = 9 or -12. Since n cannot be negative, n = 9
(14) b.
and
Step 1
On comparing the quadratic equation
= 0, with the standard f orm ax2 +
bx + c = 0, we get:
a = 1,
b=-
,
c = 60
Step 2
D = b2 - 4ac
)2 - 4(1)(60)
= (-
= 245 - 240
=5
Step 3
By using the quadratic f ormula, we get:
y=
-b ∓ √D
2a
=
∓
2×1
∴y=
or
Step 4
T hus, the roots of the quadratic equation
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= 0, are
and
.
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ID : ww-10-Quadratic-Equations [14]
(15) c. 0 and 140
Step 1
On comparing equation with standard quadratic equation ax2 + bx + c = 0, we get,
a = 5, b = -k and c = 7k.
Step 2
It is given that, the quadratic equation 5x2 – kx + 7k = 0 has two equal roots,
T heref ore, b2 - 4ac = 0
⇒ (-k)2 - 4(5)(7k) = 0
⇒ k2 - 140k = 0
⇒ k(k - 140) = 0
either, k = 0
or, k - 140 = 0
⇒ k = 140
Step 3
T hus, the value of k could be 0 or 140.
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