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A quadratic involving x and y is usually in the form 'y = ax2 + bx + c'. This form is y in terms of x, so we must rearrange it.

y = ax2 + bx + c

y/a = x2 + bx/a + c/a

y/a = x2 + bx/a + d + e, where c/a = d + e, e = (b/a)2

y/a - e = x2 + bx/a + d

y/a -e = (x + b/a)2

√(y/a - e) = x + b/a

√(y/a - e) - b/a = x

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Q: How can I solve a quadratic equations for x in terms of y?

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why will the equations x+14=37 and x-14=37 have different solutions for x

To solve for x in the equation x2 - 2x - 2 = 0, use the quadratic equation, that is: For 0 = ax2 + bx + c, the roots (values of x) are defined as: x = [-b +/- sqrt(b2 - 4ac) ] / 2a (If that's hard to understand, google "quadratic formula"). It works out to x = 2.73, -0.73.

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3x2+x-4 = 0 (3x+4)(x-1) = 0 Solutions: x = 1 and x = -4/3 By using the quadratic equation formula.

It is not possible to solve one equation in two unknowns (x and y). Two independent equations are required.

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Graph the equation then find the x intercepts.

Equations of the form z^4+az^2+a_0 are known as biquadratic equations. They are quartic equations. In general they can be solved by reducing them to a quadratic equation where x=z^2 is the variable. Then you can use the quadratic formula or factor. So plugging in x to the biquadratic giives us x^2+ax+a_0.

In general, quadratic equations have graphs that are parabolas. The quadratic formula tells us how to find the roots of a quadratic equations. If those roots are real, they are the x intercepts of the parabola.

Solve the two equations simultaneously for x. You will probably need the y value as well.Differentiate the quadratic equation.Find the value of the derivative when you substitute the value for x (from step 1) into the derivative.That is the gradient.Solve the two equations simultaneously for x. You will probably need the y value as well.Differentiate the quadratic equation.Find the value of the derivative when you substitute the value for x (from step 1) into the derivative.That is the gradient.Solve the two equations simultaneously for x. You will probably need the y value as well.Differentiate the quadratic equation.Find the value of the derivative when you substitute the value for x (from step 1) into the derivative.That is the gradient.Solve the two equations simultaneously for x. You will probably need the y value as well.Differentiate the quadratic equation.Find the value of the derivative when you substitute the value for x (from step 1) into the derivative.That is the gradient.

x^2 -8x-4=0 x=[8±√(64+16)]/2 x=(8±4√5)/2 x=4±2√5

you use difference of squaresex. X^2-4 can be factored out to (x+2)(x-2)you now have the zeros in your equation much easier

The roots of the quadratic equation are the x-intercepts of the curve.

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With great difficulty because it is not a quadratic equation or even a quadratic expression.

The quadratic formula is used all the time to solve quadratic equations, often when the factors are fractions or decimals but sometimes as the first choice of solving method. The quadratic formula is sometimes faster than completing the square or any other factoring methods. Quadratic formula find: -x-intercept -where the parabola cross the x-axis -roots -solutions

No, linear equations don't have x2. Equation with x and y are usually linear equations. Equations with either x2 or y2 (but never both) are usually quadratic equations.

Well, that's one method to solve the quadratic equation. Here is an example (using the symbol "^" for power): solve x^2 - 5x + 6 = 0 Step 1: Convert the equation to a form in which the right side is equal to zero. (Already done in this example.) Step 2: Factor the left side. In this case, (x - 3) (x - 2) = 0 Step 3: Use the fact that if a product is zero, at least one of its factors must be zero. This lets you convert the equation to two equations; x - 3 = 0 OR x - 2 = 0 Step 4: Solve each of the two equations.

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