Solve for :
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Solve for :
First, we should notice that it is a quadratic equation because there is a quadratic term (meaning raised to the second power).
The first step in solving a quadratic equation is always arranging it in a form where all terms on one side are ordered from highest to lowest power (in descending order from left to right) and 0 on the other side.
Then we can choose whether to solve it using the quadratic formula or by factoring/completing the square.
The equation in the problem is already arranged, so let's proceed with the solving technique:
We'll choose to solve it using the quadratic formula.
Let's recall it first:
The rule states that the roots of an equation in the form are .
This formula is called: "The Quadratic Formula"
Let's now solve the problem:
First, let's identify the coefficients of the terms:
Note that in the given equation there is no first-power term, so from comparing to the general form:
we can conclude that the coefficient (which is the coefficient of the first-power term in the general form) is 0.
Let's continue and get the equation's solutions (roots) by substituting the coefficients we noted earlier in the quadratic formula:
Let's continue and calculate the expression under the root and simplify the expression:
We now have a negative expression under the root and since we cannot extract a real root from a negative number, this equation has no real solutions.
In other words, there is no real value of that when substituted in the equation will give a true statement.
Therefore, the correct answer is answer D.
No solution
Solve the following equation:
\( 2x^2-8=x^2+4 \)
It means there's no real number that makes the equation true. The graph of this parabola doesn't cross the x-axis, so there are no x-intercepts.
Calculate the discriminant . If it's negative, no real solutions exist. If zero, one solution. If positive, two solutions.
Absolutely! When there's no x term, b = 0. The formula still works:
Complex solutions involve imaginary numbers (using ). For this problem, the complex solutions would be , but that's beyond basic algebra.
You can rearrange it that way! You'd get . But since no real number squared gives a negative result, there's still no real solution.
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