A mother has two children whose ages differ by 5 years. The sum of the squares of their ages is 97. The square of the​ mother's age can be found by writing the squares of the​ children's ages one after the other as a​ four-digit number. How old is the​ mother?

Respuesta :

Answer:

41 years old.

Step-by-step explanation:

Let x represent age of younger child.

We have been given that a mother has two children whose ages differ by 5 years. So the age of older child would be [tex]x+5[/tex].

The sum of the squares of their ages is 97. We can represent this information in an equation as:

[tex]x^2+(x+5)^2=97[/tex]

Let us solve for x.

[tex]x^2+x^2+10x+25=97[/tex]

[tex]2x^2+10x+25-97=0[/tex]

[tex]2x^2+10x-72=0[/tex]

Divide both sides by 2:

[tex]x^2+5x-36=0[/tex]

[tex]x^2+9x-4x-36=0[/tex]

[tex]x(x+9)-4(x+9)=0[/tex]

[tex](x+9)(x-4)=0[/tex]

[tex](x+9)=0 ; (x-4)=0[/tex]

[tex]x=-9 ; x=4[/tex]

Since age cannot be negative, therefore, age of younger child is 4 years.

Age of older child would be [tex]x+5\Rightarrow4+5=9[/tex]

Therefore, the age of older child would be 9 years.

We have been given that the square of the​ mother's age can be found by writing the squares of the​ children's ages one after the other as a​ four-digit number.

Square of 4: [tex]4^2=16[/tex]

Square of 9: [tex]9^2=81[/tex].

Square of mother's age: [tex]1681[/tex]

To find mother's age, we need to take positive square root of 1681 as:

[tex]\sqrt{1681}=41[/tex]

Therefore, the mother is 41 years old.

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