Problems on Quadratic Equation

We will solve different types of problems on quadratic equation using quadratic formula and by method of completing the squares. We know the general form of the quadratic equation i.e., ax2 + bx + c = 0, that will help us to find the nature of the roots and formation of the quadratic equation whose roots are given.

1. Solve the quadratic equation 3x2 + 6x + 2 = 0 using quadratic formula.

Solution:

The given quadratic equation is 3x2 + 6x + 2 = 0.

Now comparing the given quadratic equation with the general form of the quadratic equation ax2 + bx + c = 0 we get,

a = 3, b = 6 and c = 2

Therefore, x = βˆ’b±√b2βˆ’4ac2a

β‡’ x = βˆ’6±√62βˆ’4(3)(2)2(3)

β‡’ x = βˆ’6±√36βˆ’246

β‡’ x = βˆ’6±√126

β‡’ x = βˆ’6Β±2√36

β‡’ x = βˆ’3±√33

Hence, the given quadratic equation has two and only two roots.

The roots are βˆ’3βˆ’βˆš33 and βˆ’3βˆ’βˆš33.

 

2. Solve the equation 2x2 - 5x + 2 = 0 by the method of completing the squares.

 Solutions:

The given quadratic equation is 2x2 - 5x + 2 = 0

Now dividing both sides by 2 we get,

x2 - 52x + 1 = 0

β‡’ x2 - 52x = -1

Now adding (12Γ—βˆ’52) = 2516 on both the sides, we get

β‡’ x2 - 52x + 2516 = -1 + 2516

β‡’ (xβˆ’54)2 = 916

β‡’ (xβˆ’54)2 = (34)2

β‡’ x - 54 = Β± 34

β‡’ x = 54 Β± 34

β‡’ x = 54 - 34 and 54 + 34

β‡’ x = 24 and 84

β‡’ x = 12 and 2

Therefore, the roots of the given equation are 12 and 2.


3. Discuss the nature of the roots of the quadratic equation 4x2 - 4√3 + 3 = 0.

Solution:

The given quadratic equation is 4x2 - 4√3 + 3 = 0

Here the coefficients are real.

The discriminant D = b2 - 4ac = (-4√3 )2 - 4 βˆ™ 4 βˆ™ 3 = 48 - 48 = 0

Hence the roots of the given equation are real and equal.


4. The coefficient of x in the equation x2 + px + q = 0 was taken as 17 in place of 13 and thus its roots were found to be -2 and -15. Find the roots of the original equation.

Solution:

According to the problem -2 and -15 are the roots of the equation x2 + 17x + q = 0.

Therefore, the product of the roots = (-2)(-15) = q1

β‡’ q = 30.

Hence, the original equation is x2 – 13x + 30 = 0

β‡’ (x + 10)(x + 3) = 0

β‡’ x = -3, -10

Therefore, the roots of the original equation are -3 and -10.






11 and 12 Grade Math 

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