To solve a quadratic equation which is in the form
ax2 + bx + c = 0
we have three different ways.
(i) Factoring
(ii) Quadratic formula
(iii) Completing the square
Problem 1 :
x² - 4x + 3 = 0
Solution :
x² - 4x + 3 = 0
Since the coefficient of x2 is 1, we take the constant alone and decompose it into two parts.
Factors of 3 :
The middle term alone is negative, so both factors will be negative.
x² - x - 3x + 3 = 0
x(x - 1) - 3(x - 1) = 0
(x - 1) (x - 3) = 0
x = 1 and x = 3
So, the solution is {1, 3}.
Problem 2 :
2x² - x = 12 + x
Solution :
2x² - x = 12 + x
The given question is not in the standard form, so subtract 12 and x to get rid of it.
2x² - x - x - 12 = 0
2x² - 2x - 12 = 0
2(x² - x - 6) = 0
2(x² + 2x - 3x - 6) = 0
2(x + 2) (x - 3) = 0
(x + 2) (x - 3) = 0
x = -2 and x = 3
So, the solution is {-2, 3}.
Problem 3 :
Solve
2x2 + x - 4 = 0
Solution :
Problem 4 :
Using completing the square method, solve the equation.
x2 − 6x − 7 = 0
Solution :
x2 − 2⋅x⋅3 + 32 - 32 - 7 = 0
Write the coefficient of x as a multiple of 2.
(x - 3)2 - 32 - 7 = 0
(x - 3)2 - 9 - 7 = 0
(x - 3)2 - 16 = 0
Add 16 on both sides.
(x - 3)2 = 16
Take square roots on both sides, we get
x - 3 = √16
x - 3 = ±4
x - 3 = 4 and x - 3 = -4
x = 4 + 3 and x = - 4 + 3
x = 7 and x = -1
Problem 5 :
3(x + 7)2 + 17 = 56
Solution :
The given question is not in the standard form.
3(x + 7)2 + 17 = 56
Subtract 17 on both sides.
3(x + 7)2 + 17 - 17 = 56 - 17
3(x + 7)2 = 39
Divide by 3 on both sides.
(x + 7)2 = 39/3
(x + 7)2 = 13
(x + 7) = √13
x + 7 = ±√13
x = ±√13 - 7
Problem 6 :
2x - 5√x + 2 = 0
Solution :
It is trinomial, even though it doesn't have square, by considering the middle term (√x). Let us try to write the other terms also in terms of √x.
2(√x)2 - 5√x + 2 = 0
Let √x = t
2t2 - 5t + 2 = 0
2t2 - 4t - 1t + 2 = 0
2t(t - 2) - 1(t - 2) = 0
(2t - 1) (t - 2) = 0
Equating each factor to 0, we get
t = 1/2 and t = 2
When t = 1/2 ==> √x = 1/2 ==> x = 1/4
When t = 2 ==> √x = 2 ==> x = 4
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM