# FACTORISATION AND QUADRATIC EQUATIONQ1. Two numbers have a sum of 7 and the sum of their squares is 109. find the numbers. Q2. When a cricket ball is hit directly upwards its height above the...

FACTORISATION AND QUADRATIC EQUATION

Q1. Two numbers have a sum of 7 and the sum of their squares is 109. find the numbers.

Q2. When a cricket ball is hit directly upwards its height above the ground is given by *h=30t-5t^2 *metres, where *t *is the time after the ball is hit (in seconds). When is the ball at a height of:

a) 0 metres

b) 25 metres above the ground

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Two numbers have a sum of 7 and the sum of their squares is 109.

Let the numbers be represented by x and y.

As their sum is 7, x + y = 7. The sum of their squares is x^2 + y^2 = 109

From x + y = 7, we get x = 7 - y

Substitute in x^2 + y^2 = 109

(7 - y)^2 + y^2 = 109

49 + y^2 - 14y + y^2 = 109

2y^2 - 14y - 60 = 0

2y^2 - 20y + 6y - 60 = 0

2y(y - 10) + 6(y - 10) = 0

(2y + 6)(y - 10) = 0

y = -3, y = 10

x = 7 - y = 10 and x = -3

The two numbers are (-3, 10)

The sum of the numbers is 7.

Let them be x and 7-x

Sum the squares of the numbers = x^2+(7-x)^2 = 109

x^2+49-14x+x^2-109=0

2x^2-14x-60=0. Divide by 2 :

x^2-7x-30 = 0

(x-10)(x+3) = 0

Therefore x-10=0 or x+3=0 gives the soltion.

**x=10 **and the other number is x-10 = **-3.**

Tally:

Now 10+(-3) =7 and 10^2+(3)^2= 109

b)

Height of the ball h = 30t-5t^2

a)

When h=0,

0=30t-5^2 or

t(30-5t) = 0.

Therefore, t= 0seconds, the initial condition or

30-5t = 0 ot t= 30/5= 6 seconds, when the ball hits the ground.

b)

Whenh=25,

25=30t-5t^2. Divide by 5:

5=6t-t^2 or

t^2-6t+5=0

(t-1)(t-5)=0 . So t-1=0 or t-5=0

Therefore, t=1 seconds, while going up or

t=5 seconds when the ball is falling , itt 25 meters high.

**Q1:** Let's suppose that the 2 numbers are "a" and "b".

So a+b=7 and a^2 + b^2=109

But a^2 + b^2=(a+b)^2-2ab

So, a^2 + b^2=109 and a+b=7, that means that after subtitution: 109=7^2-2ab

109-49=-2ab

60=-2ab

ab=-30 and a+b=7+>a=7-b

We'll substitute a=7-b, into the product ab=-30

(7-b)*b=-30

7b-b^2-30=0

b^2-7b-30=0

**b1=[7+sqrt(49-120)]/2=(7+13)/2=20/2=10**

**b2=[7-sqrt(49-120)]/2=(7-13)/2=-6/2=-3**

So **a1=7-b1=7-10=-3**

**a2=7-b2=7-(-3)=7+3=10**

The system formed with the expression a+b=7, a^2+b^2=109 is known as symmetrical sustem, where the solution of the 2 unknown could be changed between them without affecting the system solution.

**Q2: ***h=30t-5t^2 *

a) When h=0, the expression above becomes *0=30t-5t^2 *

We'll multiply the equation formed with the value (-1) and we'll re-write the equation.:

5t^2-30t =0

We've noted t as common factor and we'll use this in this way:

t(5t-30)=0, knowing that a product of 2 factors which is equal to 0, one of the 2 factors is 0.

So,** t=0**, or

5t-30=0

t-6=0

**t=6**

b) When h=25, 30t-5t^2 = 25

30t-5t^2- 25=0

We'll divide the expression with 5

6t - 5t^2 - 5=0

We'll multiply the expression with (-1)

5t^2-6t+5=0,

As we can see, the expression above is always bigger than the value 0, and never 0, for any value of t. The estimation was made based on fact that delta=(-6)^2+4*5*5<0.

In terms of graphics, the expression 5t^2-6t+5=0, represents a curve with parabolic shape (due to the second grade of the equation), this curve being placed all the above of the ox axis, never intersecting it, for any values of t!