Problems on quadratic equations play a major role quantitative aptitude test. There is no competitive exam without the questions from this topic. We have already learned this topic in our lower classes.Even though we have been already taught this topic in our lower classes, we need to learn some more short cuts which are being used to solve the problems in the above topic.

The only thing we have to do is, we need to apply the appropriate short cut and solve the problems in a limited time. This limited time will be one minute or less than one minute in most of the competitive exams.

Students who are preparing to improve their aptitude skills and those who are preparing for this type of competitive test must prepare this topic in order to have better score. Because, today there is no competitive exam without questions from the topic quadratic equations problems. Whether a person is going to write placement exam to get placed or a students is going to write a competitive exam in order to get admission in university, they must be prepared to solve word problems on equations. This is the reason for why people must study this topic.

As we mentioned in the above paragraph, a person who wants to get placed in a company and a students who wants to get admission in university for higher studies must write competitive exams like placement test and entrance exam. To meet the above requirement, it is very important to score more marks in the above mentioned competitive exams. To score more marks, they have to prepare the topics like this . Preparing this topic would definitely improve their marks in the above exams. Preparing this topic is not difficult task. We are just going to remember the stuff that we have already learned in our lower classes

Students have to learn few basic operations in this topic and some additional tricks. Already we are much clear with the four basic operations which we often use in math. They are addition, subtraction, multiplication and division. Even though we are much clear with these four basic operations, we have to be knowing some more stuff to do the problems which are being asked from this topic in competitive exams. The stuff which I have mentioned above is nothing but the tricks and shortcuts which need to solve the problems in a very short time.

**Here,
we are going to have some problems on quadratic equations. You
can check your answer online and see step by step solution.**

1.The values of

2. Solve for x in the following equation.

4^{x} - 3.2^{x+2} + 2^{5} = 0

(2^{x})^{2}-3.2^{x}.2^{2} + 32 = 0

(2^{x})^{2}-12.2^{x} + 32 = 0

Taking y = 2^{x}, we get, y^{2} - 12y + 32 = 0

(y-8)(y-4)=0

y-8 = 0 or y-4 = 0

y = 8 or y = 4

2^{x} = 8 or 2^{x} = 4

2^{x} = 2^{3} or 2^{x} = 2^{2}

Therefore, x = 3 or x = 2

(2

(2

Taking y = 2

(y-8)(y-4)=0

y-8 = 0 or y-4 = 0

y = 8 or y = 4

2

2

Therefore, x = 3 or x = 2

3. If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of the reciprocals, then (b²/ac) + (bc/a²) is equal to

Let "α" and "β" be the roots of the equation ax^{2} + bx + c = 0.

Sum of the roots = Sum of the squares of their reciprocals

α+β = (1/α)^{2} + (1/β)^{2} -->
α+β=1/α^{2}+1/β^{2}

α+β=(α^{2}+β^{2})/(αβ)^{2} --> α+β=[(α+β)^{2} - 2αβ]/(αβ)^{2} ---(1)

We know, α + β = - b/a and αβ = c/a

(1)===> -b/a = [(-b/a)^{2} - 2c/a]/(c/a)^{2}

(-b/a)x(c^{2}/a^{2}) = b^{2}/a^{2} - 2ac/a^{2}

-bc^{2}/a^{3} = (b^{2} - 2ac)/a^{2}

-bc^{2} = a^{3}[(b^{2} - 2ac)/a^{2}]

-bc^{2} = a(b^{2} - 2ac)

-bc^{2} = ab^{2} - 2a^{2}c)

2a^{2}c = ab^{2} + bc^{2} (or) ab^{2} + bc^{2} = 2a^{2}c

Dividing by a^{2}c on both the sides, we get

b^{2}/ac + bc/a^{2} = 2

Sum of the roots = Sum of the squares of their reciprocals

α+β = (1/α)

α+β=(α

We know, α + β = - b/a and αβ = c/a

(1)===> -b/a = [(-b/a)

(-b/a)x(c

-bc

-bc

-bc

-bc

2a

Dividing by a

b

4. If L+M+N =0 and L M N are rationals, the roots of the equation (M+N-L)x² + (N+L-M)x + (L+M-N) = 0 are

From L+M+N = 0, we get L+M = -N, M+N = -L, N+L = -M

(M+N-L)x^{2} + (N+L-M)x + (L+M-N) = 0

(-L-L)x^{2} + (-M-M)x + (-N-N) = 0

-2Lx^{2} -2Mx - 2N = 0 ---> Lx^{2} + Mx + N = 0

In the above quadratic equation a = L, b = M and c = N

b^{2}-4ac = M^{2} - 4LN

b^{2}-4ac = (-L-N)^{2} - 4LN

b^{2}-4ac = (L+N)^{2} - 4LN

b^{2}-4ac = L^{2} + N ^{2} + 2LN - 4LN

b^{2}-4ac = L^{2} + N ^{2} - 2LN

b^{2}-4ac = (L - N)^{2}

Since b^{2}-4ac is a perfect square, the roots are real and rational

(M+N-L)x

(-L-L)x

-2Lx

In the above quadratic equation a = L, b = M and c = N

b

b

b

b

b

b

Since b

5. If p≠q and p² = 5p-3, q² = 5q-3, the equation having roots as p/q and q/p is

From p^{2} = 5p-6 and q^{2} = 5q-6, we get

p^{2}-5p+6 = 0 and q^{2}-5q+6 = 0

By solving, we get p = -2,-3 and q = -2, -3

Since p ≠ q, we can take p = -2 and q = -3

So, p/q = -2/-3 = 2/3 and q/p = -3/-2 = 3/2

Sum of the roots = p/q + q/p = 2/3 + 3/2 = 13/6

Product of the roots = (p/q)x(q/p) = (2/3)x(3/2)= 1

Formation of quadratic equation:

x^{2} - (sum of the roots)x + product of the roots = 0

x^{2} - (13/6)x + 1 =0

6x^{2} - 13x + 6 =0

Hence, the required quadratic equation is 6x^{2} - 13x + 6 =0

p

By solving, we get p = -2,-3 and q = -2, -3

Since p ≠ q, we can take p = -2 and q = -3

So, p/q = -2/-3 = 2/3 and q/p = -3/-2 = 3/2

Sum of the roots = p/q + q/p = 2/3 + 3/2 = 13/6

Product of the roots = (p/q)x(q/p) = (2/3)x(3/2)= 1

Formation of quadratic equation:

x

x

6x

Hence, the required quadratic equation is 6x

6. If one root of the equation x²-8x+m = 0 exceeds the other by 4, then the value of "m" is

In the equation x^{2} - 8x + m =0, constant term "m" is having positive sign.

So, sum of the two factors of "m" is equal to the middle term 8 and also there must be difference 4 between the two roots.

The above condition can be met, only if the two factors of "m" are 2 and 6.

Therefore m = 2x6 = 12

So, sum of the two factors of "m" is equal to the middle term 8 and also there must be difference 4 between the two roots.

The above condition can be met, only if the two factors of "m" are 2 and 6.

Therefore m = 2x6 = 12

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