Find the square
root of the number
225. I want step by step explanation
Answers
Answer:
Let the four parts be a−3d,a−d,a+d,a+3d
Sum, 4a=56
a=14
{ \: \: \: \: \mapsto\sf \dfrac{(a-3d)(a+3d)}{(a-d)(a+d)}= \dfrac{5}{6} }↦
(a−d)(a+d)
(a−3d)(a+3d)
=
6
5
Using Identity (A+B)(A-B) = A²-B²
{ \: \: \: \: \mapsto\sf \dfrac{ {a}^{2} - {(3d)}^{2} }{ {a}^{2} - {d}^{2} }= \dfrac{5}{6} }↦
a
2
−d
2
a
2
−(3d)
2
=
6
5
{ \: \: \: \: \mapsto\sf \dfrac{ {a}^{2} - 9 {d}^{2} }{ {a}^{2} - {d}^{2} }= \dfrac{5}{6} }↦
a
2
−d
2
a
2
−9d
2
=
6
5
{\: \: \: \: \mapsto\sf 6({a}^{2} - 9 {d}^{2}) = 5( {a}^{2} - {d}^{2}) }↦6(a
2
−9d
2
)=5(a
2
−d
2
)
{\: \: \: \: \mapsto\sf 6{a}^{2} - 54{d}^{2} = 5 {a}^{2} - 5{d}^{2} }↦6a
2
−54d
2
=5a
2
−5d
2
{\: \: \: \: \mapsto\sf 6{a}^{2} -5 {a}^{2} - 54{d}^{2} + 5{d}^{2}= 0 }↦6a
2
−5a
2
−54d
2
+5d
2
=0
{\: \: \: \: \mapsto\sf {a}^{2} - 49{d}^{2} = 0 }↦a
2
−49d
2
=0
Putting the value of a
{\: \: \: \: \mapsto\sf {14}^{2} - 49{d}^{2} = 0 }↦14
2
−49d
2
=0
{\: \: \: \: \mapsto\sf 196 - 49{d}^{2} = 0 }↦196−49d
2
=0
{\: \: \: \: \mapsto\sf - 49{d}^{2} = - 196 }↦−49d
2
=−196
{\: \: \: \: \mapsto\sf {d}^{2} = 196 \div 49 }↦d
2
=196÷49
{\: \: \: \: \mapsto\sf {d}^{2} = 4 }↦d
2
=4
{\: \: \: \: \mapsto\sf d = \sqrt{4} }↦d=
4
{\: \: \: \: \mapsto\sf d = \pm2 }↦d=±2
Numbers are 8,12,16,20.