Question

Two numbers when multiplied equals 45. one number is x and the other number is greater than x by 4 a) Represent the information given in equation form. b) Express as a quadratic equation c) Solve the quadratic equation to find possible values for x d) If told that the two number were positive, give the values of the two original numbers.

Answer 1

Step-by-step explanation:

a)

$x \times \frac{x}{4} = 45$

b) sorry i can't tell..

c)

$x \times \frac{x}{4} = 45 \\ = \frac{2x}{4} = 45 \\ 2x = 45 \times 4 \\ x = 180 \div 2 = 90$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

Two numbers when multiplied equals to 45.

One number is x and the other number is greater than x by 4.

So,

One number = x

and

Second number = x + 4

Now, the product of two numbers is 45.

$\bf\implies \:x(x + 4) = 45$

$\rm :\longmapsto\: {x}^{2} + 4x = 45$

$\rm :\longmapsto\: {x}^{2} + 4x - 45 = 0$

$\rm :\longmapsto\: {x}^{2} + 9x - 5x - 45 = 0$

$\rm :\longmapsto\:x(x + 9) - 5(x + 9) = 0$

$\rm :\longmapsto\:(x + 9)(x - 5)= 0$

$\rm :\longmapsto\:x = - 9 \: \: \: or \: \: \: x = 5$

But it is given that number is positive.

$\rm :\longmapsto\:x = - 9 \: \{rejected \} \: \: or \: \: \: x = 5$

$\bf\implies \:x = 5$

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:So - \begin{cases} &\sf{one \: number = 5} \\ &\sf{other \: number = 5 + 4 = 9} \end{cases}\end{gathered}\end{gathered}$

Verification :-

One number = 5

Other number = 9

So, it implies other number is greater than one number by 4.

And product of two numbers, 5 × 9 = 45

Hence, Verified