Question

Зная что x1 и x2 корни уравнения x^2-22x+49=0 найдите ответ 2x1^2+2x2^2+8x1x2

Answer 1

По т. Виета:
$x_1+x_2=22\\ x_1x_2=49$

имеем, что

$2x_1^2+2x_2^2+8x_1x_2=2(x_1^2+2x_1x_2+x_2^2-2x_1x_2)+8x_1x_2=\\ \\ =2(x_1+x_2)^2-4x_1x_2+8x_1x_2=2(x_1+x_2)^2+4x_1x_2=\\ \\ =2\cdot22^2+4\cdot49=1164$

Answer 2

$2x_1^2+2x_2^2+8x_1x_2=2(x_1^2+x_2^2+4x_1x_2)=2[(x_1+x_2)^2+2x_1x_2]=\\2[(-b)^2+2c]=2b^2+4c=2*(-22)^2+4*49=1164$

Answer 3

Solving for x1 and x2

To find the value of 2x1^2 + 2x2^2 + 8x1x2, we first need to find the values of x1 and x2, which are the roots of the equation x^2 - 22x + 49 = 0.

Using the quadratic formula, we can find the roots of the equation: x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 1, b = -22, and c = 49.

Calculating the discriminant (b^2 - 4ac): b^2 - 4ac = (-22)^2 - 4*1*49 = 484 - 196 = 288

Now, we can find the roots using the quadratic formula: x1 = (-(-22) + √288) / (2*1) = (22 + √288) / 2 x2 = (-(-22) - √288) / (2*1) = (22 - √288) / 2

Finding 2x1^2 + 2x2^2 + 8x1x2

Now that we have the values of x1 and x2, we can substitute them into the expression 2x1^2 + 2x2^2 + 8x1x2 and simplify:

2x1^2 + 2x2^2 + 8x1x2 = 2*(x1^2 + x2^2) + 8x1x2 = 2*((x1)^2 + (x2)^2) + 8x1x2 = 2*((22 + √288)^2 / 4 + (22 - √288)^2 / 4) + 8*(22 + √288)/2*(22 - √288)/2 = (22 + √288)^2 / 2 + (22 - √288)^2 / 2 + 8*(22^2 - 288) / 4 = (484 + 44√288 + 288) / 2 + (484 - 44√288 + 288) / 2 + (88 - 288) * 2 = (772 + 44√288) / 2 + (772 - 44√288) / 2 - 400 = 772 / 2 - 400 = 386 - 400 = -14

The value of 2x1^2 + 2x2^2 + 8x1x2 is -14.