40 Chapter 2
Review Exercise 2
270y.y - y.y - 3.y = 0.y 270 - y 2 - 3y =
0
= 128 ( ( ) ⎛ │ ⎝ ( ω = 128 1 2 ω 1 + ( 3 ) 2 ω 1 1 ) 4 ω 2 )
+ ( ω 3 ) 4 ω 2 ⎞ │ ⎠
∴ ω
3
=
1
- y 2
- 3y + 270 =
0
= 128 ( ω + ω 2 )
∴ 1 + ω + ω
2
=
0 y 2+ 3y - 270 =
0
= 128 ( -
1
) y 2+ 18y - 15y - 270 =
0
= -
128 y ( y + 18 ) - 15 ( y + 18 )
=
0
Q4i). Without solving the equation. Find the ( y - 15 )( y + 18 )
=
0
sum and product of the roots of
4x 2 - 1 = 0 Either or
Sol: Comparing 4x 2 + 0.x - 1 =0 y - 15 =
0
y + 18 =
0 y =
15
y = -
18 Since number of goats cannot be negative
with the quadratic equation
ax 2 + bx + c = 0 we have a = 4 , b = 0, c =
- 1 Sum of the roots So number of goats y = 15
α + β =
Putting the values of a and b Review Exercise 2 Q1. Fill in the correct circle only α + β = i). if the sum of roots of ( a + 1 ) x 2 + ( 2 a + 3 ) x + ( 3 a + 4 ) = 0 is – 1
Product of the roots
then product of roots
Putting the values of a and c O 0 O 1 O 2 O 2 ii). The sum of the roots of quadratic eq is 2 & sum of cubes of roots is 98 the eq O x 2 - 2 x - 15 = 0 O x 2 - 2 x + 15 = 0 O x 2 - 4 x + 15 = 0 O None of these iii). If abc , , positive real numbers, then both roots of eq ax 2 + bx + c = 0 are always O Imaginary O Irrational
- b- a ( )0 4 = 0 4 = 0 α . β = c a α . β = - 4 1 Q4ii). Without solving the equation. Find the sum and product of the roots of
3x 2 + 4x = 0 Sol: Comparing 3x 2 + 4x + 0 =0 with the quadratic equation
ax 2 + bx + c = 0 we have a = 3 , b = 4, c =
0 Sum of the roots O Rational O All of these iv). If a and b are the roots of 4 x 2 - 3 x + 7 = 0 then the value of 1 a + 1
b is
α + β =
- a bPutting the values of a and b α + β =- 3 4O -3/4 O 3/7 Product of the roots O -3/7 O 4/7 Q2. For what value of k the roots of the
α . β =
c a Putting the values of a and c equation 3x 2 - 5x + k =
0 are equal Sol: Comparing
3x 2 - 5x + k =0 α . β = 0 3 = 0 with the quadratic equation
ax 2 + bx + c = 0 we have a = 3 , b = - 5, c =
k
Q5. roots Find of 3x the 2 + value ( 2k + of 1 ) k x so + k that - 5 =
sum 0 is of equal the
to According to the given condition
= b 2- 4ac =
0
Discriminant
⇒ ⇒ 25 ( - 5 - ) 12k - 4 ( =
3 )( 0
k )
=
0
the Sol: product Comparing
of the 3x roots.
2 + ( 2k + 1 ) x + k - 5 = 0 with the quadratic equation
⇒ 25 =
12k
we have 3 , 2k + 1 , k - 5 According to the given condition Sum of the roots = Product of the roots i.e.,
Q3. Evaluate
Sol: As
Substituting the values of b and c
Take( - + ) + ( - - ) = ( ω ) + ( ω )
ax 2 + bx + c = 0 a = b = c =
⇒ k =
25 12 ( - 1 + i 3 ) 7 + ( - 1 -
i 3 )
7 .
b c a a by a ω = - 1 + i 2 3 and ω 2 =
- 1 - i 2 3 b c ⇒ 2 ω = - 1 + i 3 and 2 ω 2 = - 1 -
i 3 α + β =
α β - = ×
- =
- ( 2k + 1 ) = k -
5
1 i 3 7 1 i 3 7 2 7 2
2 7 ⇒ - 2k - 1 = k -
5 ⇒ + 5 - 1 = k +
2k = 2 7 ω 7 +
2
7 ω
14
⇒ 4 = 3k= 2
7 ( ω 6 ω 1 +
ω 12 ω
2
)
⇒ k =
4 3
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