How To Solve Quadratic Equations Quickly

Type 1: Solving Quadratic Equations Questions Quickly

• In each of these questions, two equations are given. You have to solve these equations and find out the values and relation of between x and y.

Question 1  Solve for the equations 17x² + 48x – 9 = 0 and 13y² – 32y + 12 = 0

Options:

A. x < y

B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

Solution     17x² + 48x – 9 = 0…. (1)

17x² + 51x – 3x – 9 = 0

(x+3) (17x – 3) = 0

Therefore, Roots of first equation are -3 and \frac{3}{17}

We know that, if sign given in the equation is + and – then their sign of roots is – and + respectively.

Therefore, the roots of the equation are – 3 and – \frac{3}{17}

Now, 13y² – 32y + 12 = 0 ……. (2)

13y² – 26y – 6y + 12 = 0

(y-2) (13y – 6) = 0

Therefore, Roots of second equation are 2 and \frac{6}{3}

We know that, If sign given in the equation is – and – then their sign of roots is + and + respectively.

Therefore, the roots of the equation are +2 and + \frac{6}{13}

Now, compare the roots – x1, + x₂, + y₁, and + y₂

It means y>x

Correct option: A

Question 2. Solve for the equations \mathbf{\sqrt{500}x =\sqrt{420}}  and \mathbf{\sqrt{260}y – \sqrt{200} = 0  }

Options

A. x < y

B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

Solution:    \sqrt{500}x -\sqrt{420} = 0 …… (1)

\sqrt{500}x =\sqrt{420} 

500 x = 420

x =\frac{420}{500}

x = \frac{210}{250}

x = 0.84

Now, \sqrt{260}y -\sqrt{200} = 0

\sqrt{260}y = \sqrt{200} 

260y = 200

y = \frac{200}{260}

y = \frac{100}{130}

y = \frac{5}{9}

y = 0.76

On comparing x and y, it is clear that x > y

Correct option: B

Question 3. Solve for the equations    x ^{2} – 11x + 24 = 0  and     2y^{2}– 9y + 9 = 0

Options

A. x < y

B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

Solution: x^{2} – 11x + 24 = 0 ……….. (1)

                 x^{2} – 8x-3x + 24 = 0

                (x – 8) (x – 3) = 0

Therefore, Roots of first equation are 8 and 3

We know that, If sign given in the equation is  – and – then their sign of roots is + and +

Therefore, the roots of the equation are +8 and + 3

Now,  2y^{2}– 9y + 9 = 0 ……. (2)

2y^{2}– 6y – 3y+ 9 = 0

(y-3) (2y – 3) = 0

Therefore, Roots of second equation are 3 and 1.5

We know that, If sign given in the equation is – and + then their sign of roots is + and +

Therefore, the roots of the equation are + 3 and + 1.5

Now, compare the roots +x₁, +x₂, + y₁, and + y₂

It means x ≥ y

Correct option: D