Question

How do I solve this?

Answer 1

Value of $x$is $- 12$.

Final Answer is $480$

Step-by-step explanation:

Assume that:

First number = x

Second number = 20 - x

Putting the given conditions,

$2x^{2} + 3(20 - x)^{2}$

Putting the formula (a-b)²=(a²+2ab+b²) in (20-x)²,

$2 {x}^{2} + 3(20^{2} - 2 \times 20 \times x + {x}^{2})$

$= > 2 {x}^{2} + 3(400 - 40x + x^{2})$

Opening the brackets,

$2 {x}^{2} + 3 \times 400 - 3 \times 40x + 3 \times {x}^{2}$

$= > 2 {x}^{2} - 1200 - 120x + 3{x}^{2}$

$= > (2x^{2} + 3 {x}^{2}) - 120x + 1200$

$= > 5 {x}^{2} - 120x + 1200$

$As \: x= - \frac{b}{2a}$

$= > - \frac{120}{2 \times 5}$

$= > - \frac{120}{10}$

$= > - 12$

Now substituting the value of x in 5x²-120x+1200,

$= > (5 \times {12}^{2} )-(120 \times 12)+1200$

$= > (5 \times 144)-(1440)+1200$

$= > 720 - 1440 + 1200$

$= > 720 - (1440 - 1200)$

$= > 720 - 240$

$= > 480$