Q&A for How to Solve Quadratic Equations

Use the square root method! If the only variable in your equation is x², move all of the other numbers to the other side of the equation. Then, find the square root of both sides of the equation.

Sorry, no. Factoring quadratics is definitely a challenge for many people. Console yourself with the knowledge that you're not likely ever to need this skill in real life. Passing your next math test is another matter.

Having a double root means that your espression is a perfect square. It also means that the discriminate is zero. Using either of these will give an equation in k that you can solve for the answer. A square takes the form (ax+b)^2= a^2 x^2 + 2abx + b^2. Obviously, then b = +1 or -1, leaving a=+/- (k+1)/2 = sqrt(k+4). Setting the discriminant of the original quadratic to zero gives (k+1)^2 - 4 (k+4) = 0. Whichever way you prefer you should get k=-3 (x^2 -2x +1 = (x-1)^2) and k=5 (9x^2 + 6x +1 = (3x+1)^2).

x²-6x+9 was factored into (x-3)(x-3) or (x-3)².

Factor into (y-18)(y-17). The solutions are y = 17 and y = 18.

Positive numbers have two square roots - the negative and positive version of each number. 0 only has one, 0. Negative numbers don't have any real square roots, but do have two complex square roots.

a*x^2 [squared] + b*x = c = 0 is the general form for any quadratic equation. If the roots of this equation are k and m [suppose], then k + m = -(b/a). & k*m = (c/a). Use these equations to solve for k and m or for x.

Remove the fraction(s) by multiplying both sides of the equation by the denominator(s) of the fraction(s). Then proceed as usual.

You cannot easily factor this expression because there does not exist two numbers whose sum is 5 and whose product is 4. We can confirm this by using the quadratic formula, where we find that x = -5/2 +- sqrt(41)/2, which are irrational numbers.

-176 +100 to each side to get rid of it. 4x>6x+352. -6x each side. -2x>352. Divide each side by -2. x>-176. Done.

The process starts when x^2-6x+9 gets factored out to (x-3)(x-3). Then, you use the FOIL method to multiply x*x, then -3x-3x to get -6x. Then, you multiply -3 by -3 to get 9.

By having the x^2 function positive, it simplifies operations later and, as happens too often with students, they forget the negative value and end up with the wrong answer. Better to set it up that way first. It also simplifies factoring. Again, it gets rid of troublesome leading negatives.

Rewrite the equation as 2x² - 7x - 15 = 0. Factor the expression: (2x + 3)(x - 5) = 0. Set each factor equal to zero to find that x = 5 and -3/2.

Subtract 2 from both sides. Then you have 3x² - 7x - 2 = 0. In this case it doesn't factor, so you will have to use the quadratic formula.

Because when the expression is set equal to zero, you can ignore the 2 (or whatever constant is factored out of the expression). It doesn't affect the value of the variable when there's a zero on the other side of the "equals" sign.

If the equation is (2x - 1)² = 1, find the square root of both sides: +/-(2x - 1) = +/-1. That means 2x - 1 = +/-1, so that 2x = 2 or 0, and x = 1 or 0. It also means that 1 - 2x = +/-1, so that -2x = 0 or -2, and x = 0 or 1. So there are three possible values of x: 1, 0 and -2.

First multiply both sides of the equation by x: x(x + 2/x) = x² + 2 = 11x . Now subtract 11x from both sides: x² - 11x + 2 = 0. This quadratic equation cannot be factored, so we would have to use the quadratic formula to solve it. See Solve Quadratic Equations Using the Quadratic Formula .

It depends on the numbers involved. With some numbers completing the square is faster. With others, the formula is faster. That's why you were taught both methods.

See Solve Simultaneous Equations Using Substitution Method , Solve Simultaneous Equations Using Elimination Method and Solve Simultaneous Equations Graphically .

The term is "discriminant."

The general form is ax² + bx + c = 0. In this case, a = 4, b = 0 and c = -20. To express it in standard form, you would write the equation as 4x² - 20 = 0.

See Section 2 of the Wikipedia article "Quadratic Formula."

Use the quadratic formula. In this case, either b or c would be zero.

This is a basic linear equation (not a quadratic). First combine terms: 3x - 6 = 0. Now add 6 to both sides: 3x = 6. Then divide both sides by 3: x = 2. See Solve Two Step Algebraic Equations .

This can be done by completing the square. Because x² + 6x + 9 is the square of (x + 3), we can add 18 to both sides of the equation to get x² + 6x + 9 = 18. Then (x + 3)² = 18. Take the square root of both sides: +/-(x + 3) = √18 = +/-3√2. This results in four possible values for x: 3 + 3√2, 3 - 3√2, -3 + 3√2 and -3 - 3√2. For further explanation, see Complete the Square .

Because that expression will not factor, you can't use the factorization method. But you can easily solve for x by using the quadratic formula (see above): 2x² - 11x + 5 = 0. +11 +/-√[(121 - 40)] / 4 = (+11 +/-√81) / 4 = (+11 +/-9) / 4 = 20/4, 2/4 = 5, ½.

In problems like this one, it's easier to solve for x if you will square (b/2) and add that to the side of the equation with the x terms in it. When you do that, you have to add the same amount to the other side in order to maintain the validity of the equation. The point of doing all this is to form a "perfect square" on the x side of the equation so that you can easily find the square root of that side and then solve for x. (In other words, you're eliminating the x² term to make it easier to find x.)

That's how completing the square works. You halve the b value and then square it.

It's there implicitly so we don't usually write it out. And you may be referring to "what if there is only one term with x, such as x^2 = 4 or -3x=0.5?." In the former, you'd just move the constant terms to other side and square root both sides, in the latter case you don't have a quadratic equation but a linear one so you'd just solve that as any other linear equation.

There are no two consecutive integers (positive or negative) whose product is 1122.