Complex Numbers and Quadratic Equations Exercise 03
Question:
Find the roots of the quadratic equation 2x^2+5x+3=0 by the method of completing the square.
Answer:
Step 1: Rewrite the equation in the form of ax^2+bx+c=0, where a, b, and c are constants.
2x^2+5x+3=0
Step 2: Divide each term by a (in this case, a = 2).
x^2 + (5/2)x + (3/2) = 0
Step 3: Rewrite the equation in the form of (x + h)^2 + k = 0.
(x + (5/4))^2 + (3/4) = 0
Step 4: Find the value of h.
h = (5/4)
Step 5: Find the value of k.
k = (3/4)
Step 6: Find the two roots of the equation.
x = (5/4) ± √((3/4))
x = (5/4) ± √(3/4)
Question:
x^2+x−2
Answer:
Step 1: Factor the expression.
(x  2)(x + 1)
Step 2: Set each factor equal to zero.
x  2 = 0 x + 1 = 0
Step 3: Solve each equation.
x = 2 x = 1
Question:
Solve : √2x^2+x+√2=0
Answer:
Step 1: Square both sides to get rid of the square root. 2x^2 + x + 2 = 0
Step 2: Rewrite the equation in standard form. 2x^2 + x  2 = 0
Step 3: Factor the equation. (2x + 1)(x  2) = 0
Step 4: Set each factor equal to zero and solve for x. 2x + 1 = 0 x = 1/2
x  2 = 0 x = 2
The solutions are x = 1/2 and x = 2.
Question:
Solve : x^2+x+1/√2=0
Answer:
Step 1: Multiply both sides of the equation by √2
x^2 + x + 1 = 0 * √2
Step 2: Simplify the left side of the equation
2x^2 + 2x + 2 = 0
Step 3: Factor the left side of the equation
2(x^2 + x + 1) = 0
Step 4: Set each factor equal to 0 and solve for x
2 = 0 OR x^2 + x + 1 = 0
Step 5: Since 2 does not equal 0, solve for x in the second equation
x^2 + x + 1 = 0
x = [1 ± √(1^2  4(1)(1))] / 2
x = [1 ± √(1  4)] / 2
x = [1 ± √(5)] / 2
x = [1 ± √5] / 2
x = [1 ± 2.236] / 2
x = 1.618 or 0.382
Question:
Solve : x2+x/√2+1=0
Answer:
Step 1: Subtract 1 from both sides of the equation to get x2 + x/√2 = 1.
Step 2: Square both sides of the equation to get x2 + 2x + 1/2 = 1.
Step 3: Subtract 1/2 from both sides of the equation to get x2 + 2x = 3/2.
Step 4: Divide both sides of the equation by 2 to get x2 + x = 3/4.
Step 5: Subtract x from both sides of the equation to get x2 = 3/4  x.
Step 6: Rewrite the equation in standard form to get x2 + x + (3/4) = 0.
Step 7: Factor the equation to get (x + (3/8))(x + (3/8)) = 0.
Step 8: Set each factor equal to 0 and solve for x to get x = 3/8.
Question:
For the quadratic equation x^2−2x+1=0, the value of x+1/x is : A −1 B 1 C 2 D −2
Answer:
Step 1: Rearrange the equation to get x^2  2x + 1 = 0
Step 2: Factor the equation to get (x  1)(x  1) = 0
Step 3: Set each factor equal to 0 and solve for x: x  1 = 0 –> x = 1
Step 4: Substitute x = 1 into x + 1/x to get 1 + 1/1 = 2
Answer: C 2
Question:
Solve the equation x^2+3x+5=0 for x.
Answer:

First, subtract 5 from both sides of the equation to isolate the x terms: x^2+3x=5.

Next, factor the left side of the equation: x(x+3)= 5.

Divide both sides of the equation by (x+3) to isolate the x term: x=5/(x+3).

To solve for x, set the denominator to 0 and solve for x: 0=x+3, x=3.

Therefore, the solutions to the equation x^2+3x+5=0 are x=3 and x=5/(x+3).
Question:
Solve : √3x^2−√2x+3√3=0
Answer:
Step 1: Square both sides of the equation: 3x^2  2x√3 + 9 = 0
Step 2: Rewrite the equation as a quadratic equation: 3x^2  2x√3 + 9 = 0 3x^2  2x√3 + 9  9 = 0  9 3x^2  2x√3 = 9
Step 3: Factor the equation: 3x^2  2x√3 = 9 3x(x  √3) = 9
Step 4: Divide both sides of the equation by 3: 3x(x  √3) = 9 (x  √3) = 3
Step 5: Solve for x: (x  √3) = 3 x  √3 = 3 x = 3 + √3
The solution to the equation is x = 3 + √3.
Question:
Solve following equation : x^2+3=0 A ±3 B ±3 C ±i3 D ±3i
Answer:
Answer: C ±i3
Question:
If α and β are the roots of the quadratic equation x^2−3x−2=0 then α/β+β/α A 3/2 B −3/2 C 13/2 D −13/2
Answer:
Step 1: Given equation is x^2−3x−2=0
Step 2: By using the quadratic formula, the roots of the equation can be determined.
α=(3+√17)/2
β=(3√17)/2
Step 3: α/β+β/α= (3+√17)/2 / (3√17)/2 + (3√17)/2 / (3+√17)/2
Step 4: Simplifying, α/β+β/α= (3+√17)/(3√17)+ (3√17)/(3+√17)
Step 5: α/β+β/α= (3+√17 + 3√17)/(3√17)(3+√17)
Step 6: α/β+β/α= 6/(1717)
Step 7: α/β+β/α= 6/0
Step 8: Since the denominator is 0, the answer is undefined.