# Solved Problems on Quadratic Equation

**Covid-19 has led the world to go through a phenomenal transition .**

**E-learning is the future today.**

**Stay Home , Stay Safe and keep learning!!!**

These problems can be solved either by splitting of middle term or by Quadratic formula.From the given word problems, we have to make a Quadratic equation and find the value accordingly.

Using these solved problems on quadratic equation, I hope you can solve your problems based on this.

**1) Divide 16 into two positive numbers such that twice the square of the larger**

part exceeds the square of the smaller part by 164.

part exceeds the square of the smaller part by 164.

**Solution :**Let the larger part be x then the smaller part will be 16 –x

Now ,

2x

^{2}= ( 16 – x )

^{2}+ 164

⇒ 2x

^{2}- ( 16 – x )

^{2}- 164 = 0

⇒ x

^{2}+ 32x – 420 = 0

⇒ ( x + 42 )( x – 10) = 0

⇒ x = - 42 or x = 10

As we want positive numbers, so x = 10 is the larger number.

Smaller number = 16 – x = 16 – 10 = 6

Hence the required numbers are 10 and 6.

----------------------------------------------------------------------

**solved problems on quadratic equation**

**2) A two digit number is such that the product of its digit is 18. When 63 is subtracted from the number, the digits interchange their places.**

Find the number.

Find the number.

**Solution :**Let the tens digit be x then the units digits = 18/x

Number = 10x + (18/x)

Number obtained by inter changing the digits = 10 (18/x) + x

∴ ( 10 x + 18/x ) – [ 10 ×(18 / x) + x ] = 63

⇒ 10 x + 18 / x – 180 / x - x = 63

& rArr; 9x – 162 / x – 63 = 0

⇒ 9x

^{2}- 162 – 63 x =0

⇒ 9x

^{2}- 63x –162 = 0

⇒ x

^{2}- 7x – 18 = 0

⇒ ( x – 9 )( x + 2) = 0

⇒ x = 9 or x = -2

But the digits never be negative.

∴ x = 9

Ten’s digit = 9

Unit digit = 18 / x = 18 / 9 = 2

So the number = 92.

----------------------------------------------------------------------

**3) Two pipes running together can fill a cistern in (40/13) minutes.**

If one pipe takes 3 minutes more than the other to fill it, find the time

taken by each pipe to fill the cistern.

If one pipe takes 3 minutes more than the other to fill it, find the time

taken by each pipe to fill the cistern.

**Solution :**Suppose, time taken by the faster pipe = x min

∴ time taken by the slower pipe = x+ 3 min

Portion of the cistern filled by the faster pipe in one minute = 1/x

⇒ Cistern filled by the faster pipe in (40/13) min

= (1 / x )× (40 / 13) = 40 / 13x

Similarly cistern filled by slower pipe in (40/13) min

= (1 / (x + 3)) (40 / 13) = 40 / 13(x + 3)

Cistern gets filled in 40/13 min

⇒ 40 / 13x + 40 / 13(x + 3) = 1

⇒ 1 / x + 1 / (x + 3) = 13 / 40 ( multiply by 13 / 40 on both sides )

⇒ [ (x + 3 ) + x ] / x (x + 3 ) = 13 / 40 [ LCD = x(x +3)]

⇒ 40 ( 2x + 3) = 13x ( x + 3) [Cross multiply ]

⇒ 80 x + 120 = 13x

^{2}+ 39x

⇒ 13x

^{2}- 41 x – 120 = 0

⇒ 13x

^{2}- 65x + 24x – 120 = 0

⇒ 13x ( x – 5) + 24 (x – 5 ) = 0

⇒ (x – 5) ( 13x + 24 ) = 0

⇒ x = 5 or x = - 24/ 13

Hence, time taken by the faster pipe = 5 min

Time taken by the slower pipe = 8 min

---------------------------------------------------------------------

Solved problems on quadratic equation

**4) Find the values of k for which the quadratic equation 5x**

^{2}- 7x + k = 0 has real and distinct roots.**Solution :**The given equation is 5x

^{2}- 7x + k = 0, which is of the form

ax

^{2}+ bx + c = 0, where,

a = 5, b = –7, c = k

We know that for real and distinct roots D ≥0

Or b

^{2}- 4ac ≥0

∴ (-7)

^{2}- 4(5)(k) ≥ 0

⇒ 49 - 20k ≥ 0

⇒ -20k ≥ -49

⇒ k ≤ 49/20

Hence, k ≤ 49 / 20 is the required value of k.

---------------------------------------------------------------------

**solved problems on quadratic equation**

**5) Find the values of k for which the quadratic equation 2x**

has real and equal roots.

^{2}- (k + 2)x + k = 0has real and equal roots.

**Solution :**The given equation is 2x

^{2}- (k + 2)x + k = 0, which is of the form

ax

^{2}+ bx + c = 0, where,

a = 2, b = –(k + 2), c = k

We know that for real and equal roots D = 0

Or b

^{2}- 4ac = 0

∴[-(k + 2)]

^{2}- 4(2)(k) = 0

⇒ k

^{2}+ 4 + 4k - 8k = 0

⇒ k

^{2}+ 4 - 4k = 0

⇒ (k - 2)

^{2}= 0

or (k - 2) = 0

or k = 2

Hence, k = 2 is the required value of k.

**Introduction of Quadratic Equations**

• Splitting of middle term

• Completing square method

• Factorization using Quadratic Formula

• Solved Problems on Quadratic Equation

• Splitting of middle term

• Completing square method

• Factorization using Quadratic Formula

• Solved Problems on Quadratic Equation

**Covid-19 has affected physical interactions between people.**

**Don't let it affect your learning.**