2X 2 4X 3 0

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2X 2 4X 3 0. Given that, 2x2 − 4x − 3 = 0. 2x2 + 4x = 0 step 2 :

Step 1 :equation at the end of step 1 : Given that, 2x2 − 4x − 3 = 0. Xem chi tiết lớp 8 toán. Take the coefficient of x , which is 2 , divide by two, giving 1 , and finally square it giving 1 Divide both sides of the equation by 2 to have 1 as the coefficient of the first term : A.2x2 + 4x + 7. X ≓ 2.269530773 x = 0 step by step solution : The calculator solution will show work to solve a quadratic equation by completing the square to solve the entered equation for real and complex roots. C.2x2 + 4x + 6. Step 3 :pulling out like terms :

∴ 2x2 − 4x = 3. To complete the square on l.h.s., we need last term = (m idl.t erm)2 4 ×f irstt erm = ( −4x)2 4 ⋅ 2x2 = 16x2 8x2 = 2 adding 2 on both sides, 2x2 − 4x +2 = 3 +2 ∴ 2(x2 −2x +1) = 5 ∴ 2(x − 1)2 = 5 ∴ (x −1)2 = 5 2 ∴ (x −1) = ± √ 5 2 ∴ x = 1 ± √ 5 2, is the desired soln. Step 1 :equation at the end of step 1 : D.2x2 + 3x + 7. Y = 2x3 + 4x2 +6x+ 12. Take the coefficient of x , which is 2 , divide by two, giving 1 , and finally square it giving 1 2x2 + 4x = 3 2 x 2 + 4 x = 3 divide each term in 2x2 + 4x = 3 2 x 2 + 4 x = 3 by 2 2 and simplify. Step 1 :equation at the end of step 1 : Step 1 :equation at the end of step 1 : Step 1 :equation at the end of step 1 : 2x2 + 4x = 0 step 2 :

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To complete the square on l.h.s., we need last term = (m idl.t erm)2 4 ×f irstt erm = ( −4x)2 4 ⋅ 2x2 = 16x2 8x2 = 2 adding 2 on both sides, 2x2 − 4x +2 = 3 +2 ∴ 2(x2 −2x +1) = 5 ∴ 2(x − 1)2 = 5 ∴ (x −1)2 = 5 2 ∴ (x −1) = ± √ 5 2 ∴ x = 1 ± √ 5 2, is the desired soln. ∴ 2x2 − 4x = 3. 2x2+14x=0 two solutions were found : Take the coefficient of x , which is 2 , divide by two, giving 1 , and finally square it giving 1 2x2 + 14x = 0 step. 2x2 + 4x = 0 step 2 : The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction. B.2x2 + 3x + 6. Web example 16 find the discriminant of the quadratic equation 2x2 4x + 3 = 0, and hence find the nature of its roots. The calculator solution will show work to solve a quadratic equation by completing the square to solve the entered equation for real and complex roots.

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Thus, the value of this polynomial never change its sign, and, for, instance, when x = 0 it has the value 3 > 0, so it always take positive values. 2x2+14x=0 two solutions were found : To complete the square on l.h.s., we need last term = (m idl.t erm)2 4 ×f irstt erm = ( −4x)2 4 ⋅ 2x2 = 16x2 8x2 = 2 adding 2 on both sides, 2x2 − 4x +2 = 3 +2 ∴ 2(x2 −2x +1) = 5 ∴ 2(x − 1)2 = 5 ∴ (x −1)2 = 5 2 ∴ (x −1) = ± √ 5 2 ∴ x = 1 ± √ 5 2, is the desired soln. Web d = ( − 4) 2 − 4 ∗ 3 ∗ 2 < 0, hence the polynomial 2 x 2 − 4 x + 3 has no real roots, and so its graph does not intersect the horizontal axis. A.2x2 + 4x + 7. 2x2 + 14x = 0 step. 2x2 + 4x = 0 step 2 : 5x2 −4x+ 3 = 0. Web example 16 find the discriminant of the quadratic equation 2x2 4x + 3 = 0, and hence find the nature of its roots. Given that, 2x2 − 4x − 3 = 0.

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Step 1 :equation at the end of step 1 : 5x2 −4x+ 3 = 0. Take the coefficient of x , which is 2 , divide by two, giving 1 , and finally square it giving 1 X ≓ 2.269530773 x = 0 step by step solution : Y = 2x3 + 4x2 +6x+ 12. Xem chi tiết lớp 8 toán câu hỏi của olm. Rearrange the equation by subtracting what is to the right of the. D.2x2 + 3x + 7. To complete the square on l.h.s., we need last term = (m idl.t erm)2 4 ×f irstt erm = ( −4x)2 4 ⋅ 2x2 = 16x2 8x2 = 2 adding 2 on both sides, 2x2 − 4x +2 = 3 +2 ∴ 2(x2 −2x +1) = 5 ∴ 2(x − 1)2 = 5 ∴ (x −1)2 = 5 2 ∴ (x −1) = ± √ 5 2 ∴ x = 1 ± √ 5 2, is the desired soln. Web example 16 find the discriminant of the quadratic equation 2x2 4x + 3 = 0, and hence find the nature of its roots.

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Web (x2 + 3)(2x +4) or (x+i 3)(x −i 3)(2x +4) explanation: 2x2 4x + 3 = 0 comparing equation with ax2 + bx + c = 0 a = 2, b = 4, c = 3 we know that d = b2 4ac d = ( 4)2 4 (2) (3) d = 16 24 d = 8 since d <<strong>0</strong> so, the given equation has no real roots. Take the coefficient of x , which is 2 , divide by two, giving 1 , and finally square it giving 1 Given that, 2x2 − 4x − 3 = 0. 2x2 + 4x = 3 2 x 2 + 4 x = 3 divide each term in 2x2 + 4x = 3 2 x 2 + 4 x = 3 by 2 2 and simplify. Step 1 :equation at the end of step 1 : B.2x2 + 3x + 6. Step 1 :equation at the end of step 1 : Web 2x2+4x=0 two solutions were found : Web d = ( − 4) 2 − 4 ∗ 3 ∗ 2 < 0, hence the polynomial 2 x 2 − 4 x + 3 has no real roots, and so its graph does not intersect the horizontal axis.

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