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I suspect that the answer to the question is NO.
There are no such real numbers.Let the numbers be x and (2-x). The equation representing the question is:x(2-x) = 24 2x-x2 = 24 2x = x2+24 0 = x2-2x+24 x2-2x+24=0.In solving this equation, a=1, b=-2 and c=24. The discriminant is b2-4ac = (-2)2-4(1)(24) = 4-96 = -92.Since the discriminant is negative, the equation has no solution.
Derivation of x2 or 2x is 2.
x2 + 2x = 3 x2 + 2x - 3 = 0 x2 + 2x - 3 = y Graph the equation in a calculator and find its zeros, which are x = -3 and x = 1. Check: x2 + 2x = 3 (-3)2 + 2(-3) =? 3 9 - 6 =? 3 3 = 3 True (1)2 + 2(1) =? 3 1 + 2 =? 3 3 = 3 True
If x2 + 5 = -2x, then x2 + 2x + 5 = 0 This equation cannot be readily factored so using the quadratic formula :- x= {-2 ± √[(-2)2 - 4x5]} ÷ 2 = {-2 ±√-16} ÷ 2 = -1 ± 2i Therefore x = -1 + 2i or x = -1 - 2i
y = x2 - 2x + 10 This is a minimum when dy/dx = 0 dy/dx = 2x - 2 : When 2x - 2 = 0 then 2x = 2 : x = 1 Substituting in the equation gives y = 12 - (2 x 1) + 10 = 1 - 2 + 10 = 9.
(x2+3x-4) (x-4)
x2 + 2x - 7 is a quadratic polynomial. Every polynomial defines a function called P. Any value of x for which P(x) = 0 is a root of the equation and a zero of the function. x2 + 2x - 7 = 0 add 1 and 7 to both sides x2 + 2x + 1 = 8 (x + 1)2 = 8 take the square root of both sides x + 1 = ± √8 subtract 1 to both sides x = -1 ± 2√2 Thus, the roots of the equation are -1 - 2√2 and -1 + 2√2.
x2 + 2x + 5 = 0x2 + 2x + 5 - 5 = 0 - 5x2 + 2x = -5x2 + 2x + (2/2)2 = -5 + (2/2)2x2 + 2x + 12 = -5 + 1(x + 1)2 = -4sq. root of (x + 1)2 = sq. root of -4|x + 1| = 2ix + 1 = +&- 2ix + 1 - 1 = -1 +&- 2ix = -1 +&- 2i
x²+2x-15=0 x1=-2/2 - Square root of ((2/2)²+15) x1=-1-4 x1=-5 x2=-2/2 + Square root of ((2/2)²+15) x2=-1+4 x2=3
There are no such real numbers.Let the numbers be x and (2-x). The equation representing the question is:x(2-x) = 24 2x-x2 = 24 2x = x2+24 0 = x2-2x+24 x2-2x+24=0.In solving this equation, a=1, b=-2 and c=24. The discriminant is b2-4ac = (-2)2-4(1)(24) = 4-96 = -92.Since the discriminant is negative, the equation has no solution.
Yes. For example, the equation x2 = 2, which in standard form is x2 - 2 = 0, has the two solutions x = square root of 2, and x = minus square root of 2.
It is a quadratic equation with no real roots or real solutions. In the complex domain, the solutions are 1 +/- i where i is the imaginary square root of -1.
x2 + 2x + 2 = 0 Use the quadratic formula: x = (-2 +- sqrt(4 - 8))/2 x = (-2 +- 2i)/2 = -1 +- i
Derivation of x2 or 2x is 2.
x2 + 2x - 7 is a quadratic polynomial. Every polynomial defines a function called P. Any value of x for which P(x) = 0 is a root of the equation and a zero of the function. x2 + 2x - 7 = 0 add 1 and 7 to both sides x2 + 2x + 1 = 8 (x + 1)2 = 8 take the square root of both sides x + 1 = ± √8 subtract 1 to both sides x = -1 ± 2√2 Thus, the roots of the equation are -1 - 2√2 and -1 + 2√2.
x2 + 2x = 3 x2 + 2x - 3 = 0 x2 + 2x - 3 = y Graph the equation in a calculator and find its zeros, which are x = -3 and x = 1. Check: x2 + 2x = 3 (-3)2 + 2(-3) =? 3 9 - 6 =? 3 3 = 3 True (1)2 + 2(1) =? 3 1 + 2 =? 3 3 = 3 True
If x2 + 5 = -2x, then x2 + 2x + 5 = 0 This equation cannot be readily factored so using the quadratic formula :- x= {-2 ± √[(-2)2 - 4x5]} ÷ 2 = {-2 ±√-16} ÷ 2 = -1 ± 2i Therefore x = -1 + 2i or x = -1 - 2i