The polynomial which is in the form of ax^{2} + bx + c is known as quadratic polynomial. Here b or c may be 0 some times.
There are two types,
If the leading coefficient is (a = 1), then take the constant c and decompose into two terms.
If the leading coefficient is (a is not 1), then take a and c and multiply it.
Such that the product of those two factors must be equal to constant and when we simplify those two values, we should get the middle term (b).
Finally we may get factors using grouping. After getting factors, equating each factors to zero, we can find the value of x.
Solve the quadratic equations.
Problem 1 :
x^{2} + 11x + 28 = 0
a. x = -7and -4 c. x = -4 and 7
b. x = -7 and 4 d. x = 4 and 7
Solution :
x^{2} + 11x + 28 = 0
Leading coefficient = 1.
x^{2} + 4x + 7x + 28 = 0
x(x + 4) + 7(x + 4) = 0
(x + 7) (x + 4) = 0
x + 7 = 0 x = -7 |
x + 4 = 0 x = -4 |
So, option (a) is correct.
Problem 2 :
Factories the quadratic polynomial
6x^{2} - 4x + 8
a. 6x(3x – 2) c. 6x^{2} – 4x + 8
b. 3(2x^{2} – 4x + 8) d. 2(3x^{2} – 2x + 4)
Solution :
6x^{2} - 4x + 8
= 2(3x^{2} – 2x + 4)
So, option (d) is correct.
Problem 3 :
-12x^{2} - 8x = 0
a. x = 0 and x = 12 c. x = 0 and x = -2
b. x = 0 and x = -2/3 d. x = 0 and x = -3/2
Solution :
-12x^{2} - 8x = 0
-4x(3x + 2) = 0
-4x = 0 and 3x + 2 = 0
x = 0 and x = -2/3
So, option (b) is correct.
Problem 4 :
Solve by factoring.
4x^{2} + 10x – 24 = 0
a. 3/2, -1 c. 4, -1
b. -4, 3/2 d. -4, 4
Solution :
4x^{2} + 10x – 24 = 0
2(2x^{2} + 5x – 12) = 0
Dividing 2 on both sides.
2x^{2} + 5x – 12 = 0
By using algebraic expression
2x^{2} + 8x – 3x – 12 = 0
2x(x + 4) – 3(x + 4) = 0
(2x – 3) (x + 4) = 0
2x – 3 = 0 and x + 4 = 0
2x = 3
x = 3/2 , x = -4
So, option (b) is correct.
Problem 5 :
6x^{2} = 42
a. -√7, √42 c. √7, -√7
b. -√42/6, √42/6 d. √7
Solution :
6x^{2} = 42
x^{2} = 42/6
x^{2} = 7
x = ±√7
So, option (c) is correct.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM