# SOLVING 2nd DEGREE ALGEBRAIC EQUATIONS

As we have seen in post ( SOLVING 1stDEGREE ALGEBRAIC EQUATIONS ), a degree is the maximum power of unknown present in the equation. Hence the 2nd degree algebraic equation will have form of ax^2 + bx + c = 0. The value of a, b and c could be any real value, it could be 0, negative or positive. All of the below equation represent the 2nd degree algebraic equation –

x^2 + x + 6 = 0

x^2 + 3x – 10 = 0

x^2 + x = 0

x^2 + 9 = 0

To find the solution of any algebraic equation is to find the root of the equation. Root of the equation is the value that satisfies the equation. The degree of the equation provides the maximum number of root (possible solution) possible for the equation. Hence, any 2nd degree expression can have maximum two roots.

Sum of roots ->

ax^2 + bx + c = 0

Suppose the provided equation has two roots x1 and x2. The value of sum of these two roots (x1 + x2) will be,

(x1 + x2) = -b/a

For e.g.

x^2 + 3x – 10 = 0

In above equation,

a = 1

b = 3

c = -10

Suppose the above equation has two roots x1 and x2 then,

X1 + x2 = -b/a = -3/1 = -3

Product of roots ->

ax^2 + bx + c = 0

Suppose the provided equation has two roots x1 and x2. The value of product of these two roots (x1 * x2) will be,

(x1 * x2) = c/a

For e.g.

x^2 + 3x – 10 = 0

In above equation,

a = 1

b = 3

c = -10

Suppose the above equation has two roots x1 and x2 then,

x1 * x2 = c/a = -10/1 = -10

Solving Equation –

Now let’s solve the equation,

x^2 + 3x – 10 = 0

To solve the 2nd degree equation, try to separate the middle term in such way that the sum of the separated term is equal to the middle term and the product of the separated term equals to the product of first and last term.

In above equation, we can separate 3x as (5x – 2x). The (5x – 2x) will be equal to 3x and product of 5x and -2x will be -10x.

x^2 + 5x – 2x – 10 = 0

Now, take the common –

x (x + 5) – 2 (x + 5) = 0

(x-2) * (x + 5) = 0

Now the roots will be

x – 2 = 0         or        x + 5 = 0

x = 2               or        x = -5

x = 2 or -5

Sum of roots = 2 -5 = -3

Sum of roots = -b/a

Product of roots = 2 * -5 = -10

Product of roots = c/a

Now, we will solve GMAT question on similar topic –

Question

Which of the following equations is NOT equivalent to 10y^2=(x+2)(x−2)?

(A) 30y^2=3x^2-12
(B) 20y^2=(2x-4)(x+2)
(C) 10y^2+4=x^2
(D) 5y^2=x^2-2
(E) y^2=(x^2-4)/10

Solution –

To solve the given question, first simplify the equation

10y^2=(x+2)(x−2)                                        ……(1)

10y^2=(x+2)(x−2)

=(x)(x−2) +2(x−2)

= x^2 – 2x + 2x – 4

= x^2 – 4

You can also use the below property –

(x + a) (x – a) = (x^2 – a^2)

10y^2=(x+2)(x−2) = x^2 – 2^2 = x^2 – 4

10y^2 = x^2 – 4 …………(2)

As all the given option has y on left side we’ll keep the same as well.

Let’s analyze the given option,

(A) 30y^2=3x^2-12

30y^2=3x^2-12

Dividing above equation by 3, we get

10y^2 = x^2 – 4

This equation matches with equation (2), hence this is a wrong option.

(B) 20y^2=(2x-4)(x+2)

Dividing above equation by 2, we get

10y^2=(x-2)(x+2)

This equation matches with equation (1), hence this is a wrong option.

(C) 10y^2+4=x^2
Adding -4 on both side of the equation, we get

10y^2+4 – 4 =x^2 – 4

10y^2 =x^2 – 4

This is the equation matched with equation (2), hence this is a wrong option.

(D) 5y^2=x^2-2

Multiply by 2 on both sides, we get

10y^2=2x^2-4

This equation has one extra x^2 compare to equation (2), hence it does not matches with the given answer. This is the right option.

(E) y^2=(x^2-4)/10

Multiply by 10 on both sides, we get

10y^2=(x^2-4)

This equation matches with equation (2), hence this is a wrong option.

The correct option for the given question is D.