... 2. Solve the equation + 12a + 34x2 - 12x - 35 = 0 . The equation may be written ( x + 12x + 36x2 ) -2x - 12x = 35 ... square root , ( x2 + 6x ) − 1 = ± 6 . x2 + 6x = 7 or — 5 . - Completing the square , x2 + 6x + 9 = 16 or 4 . Extracting the ...

... square root is 5 x2 - 3x + 4 . TEST . In the ... 2 + 2 ( 5 x ) ( 4 ) = 49 x2 , showing the given polynomial to be a perfect square . 2. 49 +28 a3 - 38 x2-12x + 9 . 3. 36 x6-36 x3 − 3 x1 + 6 ... square , first SQUARE ROOT OF POLYNOMIALS 105.

... ( 2–52 ) . Hence , we have the Equation ( –5 ) +10 ( –5 ) = − 24 . Completing the square , ( x2−5 ) +10 ( x2−5x ) + 25 = 1 . Extracting the square root , x2-5x + 5 = ± 1 ... ( x2 — 4ax ) 6 ( 182 ) 81 SOLUTION OF CERTAIN BIQUADRATICS .

... x = 2 ; in the second example , r2 ab , so that x = ± √ ab . - Problem ... square root . An affected quadratic equation is a quadratic equation ... 6 X 5 , and 12 + 5 = 6 X 1 . ( 2 ) COMPLETING THE SQUARE Problem . Solve the ...

... x2 + 2ry + y2 = 36 Pit ( 1 ) + y2 = 20 Subtract , 2.xy = 16 Subtract ( 4 ) from ( 1 ) , x2 - 2xy + y2 = 4 Extract square root , But ( 2 ) Therefore , adding , x - y = 12 x + y = 2x 8 or 4 Hence , Substitute in ( 2 ) , 2 . { x + y = 6 . xy = ...

... 2 9x + 6x2 - 2x - 12x3 + 16x2 - 12x3 + 4x2 6.x2 - 4x + 2 / 12x2 8x + 4 12x2 8x + 4 GYM * * 2. 4x + 12x5 + 5x1 - 2x3 ... square root of 4.r1 — 16x3 + 24x2 — 16x + 4 . Ans . 2x2 4x + 2 . -- 4. Extract the square root of 16x1 + 24x3 + ...

... x - 5 = 0 , then x = 5. Quadratic 1 + - = 1 1 30 y a 1 + - = " 1 1 1 + = ย b y 2 ... square root of each side be taken , we immediately ob- tain the value of ... 6 . Hence = 6-1 , that is , x = +5 , or x = - 7 , and either 5 × 5 + 2 x 5 ...

... x 28 x2 - 16 x + 16 . - x3 +28 Solution . Square root = Given square = 2 a a2 = ( 3x2 ) 2 = a + b + c 3 x2 2x + 4 9x4 12 x 28 x2 . -- 16 x + 16 9x4 2.3x2 = 6 ... square root is exact . Example 2. Find the square root of 16 x6 24 x1 +25 x1 ...

... 2 2 e1 = = √ 6 x 2 √ 10 4 ( x2 − ) 3 13 e2 = from sympy import symbols, integrate, sqrt x = symbols('x') f0, f1, f2 = 1, x, x**2 # function to integrate two functions def inner_prod(g1,g2): e0 = f0/sqrt(inner_prod(f0,f0)) return integrate( ...

... x + 16 . x3 Solution . Square root = Given square = a2 = ( 3x2 ) 2 = α + b + c 3x2 - 2x + 4 9x4 12x8 +28 x2 - 16x + 16 - 2 a 2.3x2 = 6 ... square root is exact . Example 2 . Find the square root of 16 26-24x5 + 25 x4 - 52 x3 +34 x2 - 20 x ...