Question

QUADRATIC EQUATIONS
Example 11 : Find two consecutive odd positive integers, sum of whose squares
is 290.
integers be x. Then, the​

Answer 1

Question :-

Find two consecutive odd positive integers, sum of whose squares is 290.

$\large\underline{\sf{Solution-}}$

We know, any two consecutive odd positive integers are differ by 2.

So, Let assume that

First odd positive integer = x

Second odd positive integer = x + 2

According to statement,

Sum of squares of these two numbers is 290.

$\rm \: {x}^{2} + {(x + 2)}^{2} = 290$

$\rm \: {x}^{2} + {x}^{2} + {2}^{2} + 2 \times 2 \times x = 290$

$\rm \: 2{x}^{2} + 4 + 4x = 290$

$\rm \: 2{x}^{2} + 4x + 4 - 290 = 0$

$\rm \: 2{x}^{2} + 4x - 286 = 0$

$\rm \: 2({x}^{2} + 2x - 143) = 0$

$\rm \: {x}^{2} + 2x - 143 = 0$

$\rm \: {x}^{2} + 13x - 11x - 143 = 0$

$\rm \: x(x + 13) - 11(x + 13) = 0$

$\rm \: (x + 13)(x - 11) = 0$

$\rm\implies \:x \: = - \: 13 \: \{rejected \} \: \: or \: \: x \: = \: 11$

So,

First odd positive integer, x = 11

Second odd positive integer, x + 2 = 11 + 2 = 13

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Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac