Question

The difference of square of two numbers is 180 the square of the smaller number is 8 times the larger number Find the two number?

Answer 1

$\large\bf\underline{Given:-}$

The difference of squares of two numbers is 180.

square of smaller number is 8 times the larger number.

$\large\bf\underline {To \: find:-}$

Both numbers.

$\huge\bf\underline{Solution:-}$

◈ Let the larger number be x

◈ Let the smaller number be y

$\large \: \: \blacktriangleright \underline{ \bf According \: to \: question: - }$

The square of the smaller number is 8 times the larger number.

$\rm \rightarrowtail \: {y}^{2} = 8x........(i)$

The difference of square of two numbers is 180.

$\rm \rightarrowtail \: {x}^{2} - {y}^{2} = 180 \: ........(ii)$

Substituting value of y² = 8x from (i) in (ii)

$\rm \rightarrowtail \: {x}^{2} - 8x = 180$

So degree of the equation is 2 so it's a quadratic equation.

Now,

» Solving the quadratic equation by factorisation method.

$\rm \rightarrowtail \: {x}^{2} - 8x - 180 = 0 \\ \\ \rm \rightarrowtail \: {x}^{2} - 18x + 10x - 180 \\ \\ \rm \rightarrowtail \:x(x - 18) + 10(x - 18) \\ \\ \rm \rightarrowtail \:(x + 10)(x - 18) \\ \\ \bf \rightarrowtail \:x = - 10 \: or \: x = 18$

When , x = -10 then value of y is :-

$\rm \rightarrowtail \: {y}^{2} = 8x \\ \\ \rm \rightarrowtail \: {y}^{2} = 8 \times (- 10) \\ \\ \rm \rightarrowtail \: {y}^{2} = - 80 \\ \\ \rm \rightarrowtail \:y = \pm\sqrt{ - 80}$

So we can't take the negative number in root so the value of y = √-80 is not possible.

Now , when x = 18 then value of y :-

$\rm \rightarrowtail \: {y}^{2} = 8x \\ \\ \rm \rightarrowtail \: {y}^{2} = 8 \times 18 \\ \\ \rm \rightarrowtail \: {y}^{2} = 144 \\ \\ \rm \rightarrowtail \:y = \pm\sqrt{144} \\ \\ \rm \rightarrowtail \:y = \pm12$

Now ,

The two numbers are :-

• ≫ Smaller number = 12 or -12
• ≫ Larger number = 18

Answer 2

$\sf{\underline{Answer\::}}$

Larger Number,

• y = 18

Smaller Number,

• x = -12 or 12

$\sf{\underline{Explanation\::}}$

The difference of square of two random numbers is 180. The square of the smaller number of the two random number is 8 times the larger number.

We need to figure out which are those two random numbers satisfying the condition in our provided problem.

Let's have some assumptions to make things simpler.

Let's tag the smaller number as x & the larger number as y.

So,the difference between square of y and x is 180.

Form the 1st equation,

• x² - y² = 180

Now, the next thing says, square of x is 8 times y.

Forming our next equation,

• x² = 8y

So,we can clearly see that the value of x² from the above equation can be used in our 1st equation.

=> $\sf{8y\:-\:y^2\:=180}$

=> $\sf{-y^2\:+8y\:=180}$

=> $\sf{-y^2\:+8y-180=0}$

=> $\sf{y^2\:-8y+180=0}$

Clearly the above equation is in the general form of a quadratic equation, so solve for y using middle term splitting method.

=> $\sf{y^2\:-18y\:+\:10y-180=0}$

=> $\sf{y(y-18)\:+10(y-18)=0}$

=> $\sf{(y-18)\:\:\:(y+10)=0}$

=> $\sf{y-18=0\:\:\:y+10=0}$

=> $\sf{y=18\:\:\:or\:\:y=-10}$

So,the value of the greater number y = -10 or 18.

So this gives way to two possibilities which are,

• When y = -10
• When y = 18

Substitute the two values of y in the 2nd equation to get the value of smaller number.

=> $\sf{x^2\:=8y}$

=> $\sf{x^2\:=8(-10)}$

=> $\sf{x^2\:=-80}$

=> $\sf{x\:=\:\sqrt{-80}}$

=> $\sf{x\:=\:\pm\:\sqrt{-80}}$

Now,we know that we can't have negative number in roots. So,the possibility of value of y = -10 is not possible.

Now, using the value of y = 18 in the second equation.

=> $\sf{x^2\:=8y}$

=> $\sf{x^2\:=8(18)}$

=> $\sf{x^2\:=144}$

=> $\sf{x\:=\:\sqrt{144}}$

=> $\sf{x\:=12\:\:or\:\:-12}$

So,the two random number pair of larger and smaller number based on the given question,

1. x = 12 & y = 18
2. x = - 12 & y = 18