The following are some solved problems on quadratic equation.
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.

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.
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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.

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.
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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.

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
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4) Find the values of k for which the quadratic equation 5x2 - 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.
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5) Find the values of k for which the quadratic equation 2x2 - (k + 2)x + k = 0
has 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.