Question

find two positive numbers such that sum of three times the square of 1 and the 4 times the square of other is 43 and the difference between the square is 5​

Answer 1

♠let the 1st no. be x
and 2nd no. be y.

• Case -1

3(x)²+4(y)² = 43
3x² +4y² =43. ....i)

• Case -2

(x)²-(y)² = 5
x²- y² =5. ..........ii)

♣Solve i) and ii) eq. ,, you will get the answer!!

★★Try to help yourself by own !!!! mostly!!

♥ Hope it will help you!!!

Answer 2

Let 1st no. be X

Let 2nd no. be Y

$case \: \: \: 1 \: )\: \: \: \: 3 {x}^{2} + 4 {y}^{2} = 43$

$case \: \: \: \: 2 )\: \: \: {x}^{2} - {y}^{2} = 5$

$eleminating \: \: \: them \:$

$3 {x}^{2} + 4 {y}^{2} = 43 \\ {x}^{2} - {y}^{2} = 5$

$3 {x}^{2} + 4 {y}^{2} = 43 \\ 3 {x}^{2} - 3 {y}^{2} = 5$

${y}^{2} = 38$

$y = \sqrt{38}$

$put \: this \: value \: in \: \: 2$

${x}^{2} - 38 \: = 5$

${x}^{2} = 43$

$x = \sqrt{43}$

two no. are this one

$x = \sqrt{43} \\ y = \sqrt{38}$

hope this answer is helpful.