Question

The sum of 2 positive numbers is 151. the lesser number is 19 more than the square root of the greater number. what is the value of the greater number minus the lesser number?

Answer 1

Answer:

Let the two positive number be x and y.

Assume $x>y$

From the given condition: The sum of 2 positive number is 151 and the lesser number is 19 more than the square root of the greater number.

then we have,  

$x+y=151$     .....[1]

$y= 19+\sqrt{x}$

Substitute the value of $y= 19+\sqrt{x}$ in equation [1],

$x+19+\sqrt{x} =151$

Subtract 19 from both the sides, we get

$x+\sqrt{x}+19-19=151-19$

on simplify:

$x+\sqrt{x} = 132$ or

$\sqrt{x} =132-x$

squaring both the sides, we get

$(\sqrt{x} )^2=(132-x)^2$

Using Identities  $(a-b)^2=a^2+b^2-2ab$ on right hand side of above expression:

$x=132^2 +x^2-2\cdot 132\cdot x$ or

$x= 17424 +x^2-264x$ or we can write it as:

$x^2-265x+17424 =0$

By, solving above quadratic equation we get,

x =121

to find the value of y:

$x+y=151$

Substitute the value of x in above equation to get value for y:

$121+y=151$

⇒ y= 30

The value of greatest number minus the  lesser would be x-y, i.e,  121-30 = 91

Answer 2

The value of the greater number minus the lesser number is $\boxed{191}$.

Further Explanation:

Let the greater positive number be $\boxed{x}$.

Let the lesser positive number be $\boxed{y}$.

The first condition is that the sum of $2$ positive numbers is $151$.

The equation from the first condition can be expressed as,

$\boxed{x + y = 151}$

The second condition is that the lesser number is $19$ more than the square root of the greater number.

$\boxed{y = 19 + \sqrt x }$

Substitute the value $y = 19 + \sqrt x$ in equation $x + y = 151$ to obtain the value of $x$.

$\begin{aligned} x +\left( {19+\sqrt x }\right)&= 151 \\ x + \sqrt x &= 151 - 19 \\ x + \sqrt x &= 132 \\ \sqrt x &= 132 - x \\ \end{aligned}$

Square both the sides.

$\boxed{{\left( {\sqrt x } \right)^2} = {\left( {132 - x} \right)^2}}$

Use the identity $\boxed{{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab}$.

$\begin{aligned} x &= {132^2} + {x^2} - 2 \times 132 \times x \\ x &= {132^2} + {x^2} - 264x \\ 0 &= {x^2} - 265x + 17424 \\\end{aligned}$

Solve the above quadratic equation to obtain the value of $x$.

$\begin{aligned} {x^2} - 121x - 144x + 17424 &= 0 \\ x\left( {x - 121} \right) - 144\left( {x - 121} \right) &= 0 \\ \left( {x - 121} \right) \times \left( {x - 144} \right) &= 0 \\\end{aligned}$

The value of $x$ can be $121$ and $144$.

The value of $x=144$ cannot be possible as it doesn’t satisfies the two condition. Therefore the value of $x$ is $\boxed{121}$.

The greater number is $\boxed{121}$.

Substitute $121$ for $x$ in equation $x + y = 151$ to obtain the value of $y$.

$\begin{aligned}121 + y &= 151 \\y &= 151 - 121 \\ y &= 30 \\\end{aligned}$

The difference between the greater number and smaller number can be obtained as,

$\begin{aligned}\\\text{Difference} &= x - y \\ &= 121 - 30 \\ &= 91 \\\end{aligned}$

The value of the greater number minus the lesser number is $\boxed{191}$.

Learn more:

1. Learn more about unit conversion brainly.com/question/4837736

2. Learn more about non-collinear brainly.com/question/4165000

Answer details:

Grade: Middle School

Subject: Mathematics

Chapter: Quadratic Equations

Keywords: sum, positive number, square root, lesser number, greater, number, quadratic equation, minus, sum of 2 positive numbers is 151, difference, lesser number is 19 more than the square root of greater, number, value.