Question

hey guys pls solve quadratic equation​

Answer 1

4) let larger number be x

and smaller number be y

According to problem,

${x}^{2} + {y}^{2} = 208....eq(1)$

and also given that ,

square of larger number is 18 times the smaller number.

So,

${x}^{2} = 18y....eq(2)$

substituting the values , we get

${y}^{2} + 18y = 208$

${y}^{2} + 18y - 208 = 0$

${y}^{2} + 26y - 8y - 208 = 0$

$y(y + 26) - 8(y + 26) = 0$

$(y + 26)(y - 8) = 0$

$y = - 26 \: \: or \: \: 8$

We know that,number can not be negative.

So,y = 8

Substitute y = 8 in eq (2) , we get

${x}^{2} = 18(8)$

${x}^{2} = 144$

$x = \sqrt{144}$

$so \: x = 12$

• the larger number = 12 cm
• the smaller number = 8 cm