Expert Answer to "The sum of the squares of two negative..."

High School
Algebra
Pre-Algebra
Negative numbers and coordinate plane

abondantQ

Answered question

2021-01-23

The sum of the squares of two negative numbers is 106 and the difference of the squares of the number is 56 find the numbers

Answer & Explanation

Clara Reese

Skilled2021-01-24Added 120 answers

Calculation:
Let two negative numbers are x and y.
The sum of the square of two negative numbers is 106.
$x^{2} + y^{2} = 106$ (1)
And the difference of the square of the number is 56.
$x^{2} - y^{2} = 56$ (2)
Add equation (1) and (2).
$(x^{2} + y^{2}) + (x^{2} - y^{2}) = 106 + 56$
$2 x^{2} + 0 = 162$
$2 x^{2} = 162$
$(2 x^{2}) / 2 = 162 / 2$ [Divide by 2]
$x^{2} = 81$
Taking square root.
$\sqrt{x^{2}} = \pm \sqrt{81}$
$x = \pm \sqrt{9^{2}}$
$x = \pm 9$
x=-9, because x is negative
Substitute x = -9 in equation (1).
$- 9^{2} + y^{2} = 106$
$81 + y^{2} = 106$
$- 81 + 81 + y^{2} = 106 - 81$ [Substract 81 from both sides]
$= y^{2} = 25$
Taking square root.
$\sqrt{y^{2}} = \pm \sqrt{25}$
$y = \sqrt{5^{2}}$
$y = \pm 5$
y=-5, because y is negative
Thus, the negative numbers are x = -9 and y = -5.

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