Solve 2x²+6x-7 = x²+2x-2: Finding the Value of X

Q: How do I know which two numbers to use for factoring?

Look for two numbers that multiply to give the constant term (-5) and add to give the middle coefficient (4). For $ x^2+4x-5 $, you need 5 × (-1) = -5 and 5 + (-1) = 4.

Q: What if the quadratic doesn't factor easily?

If you can't find two numbers that work, try the quadratic formula: $ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $. This works for any quadratic equation!

Q: Why do I get two answers for x?

Quadratic equations usually have two solutions because a parabola can cross the x-axis at two points. When $ (x+5)(x-1) = 0 $, either factor can equal zero.

Q: How do I check if both solutions are correct?

Substitute each value back into the original equation. For x = -5: both sides equal 13. For x = 1: both sides equal 1. If both check out, you have the right answers!

Q: Do I always need to get zero on one side?

Yes! For factoring to work, you need the equation in the form $ ax^2 + bx + c = 0 $. This is called standard form and is essential for the factoring method.

Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find X

00:03 Arrange the equation so the right side equals 0

00:19 Group terms

00:33 Pay attention to the trinomial coefficients

00:38 We want to find 2 numbers

00:48 whose sum equals B and product equals C

00:52 These are the appropriate numbers

00:57 Therefore these are the numbers we'll put in parentheses

01:03 Find the solutions that zero each factor

01:07 Isolate X, this is one solution

01:16 Isolate X, this is the second solution

01:26 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

^{$2x^2+6x-7=x^2+2x-2$}

Solve the following problem:

2

Step-by-step solution

Let's solve the given equation:

$2x^2+6x-7=x^2+2x-2$

Let's begin by rearranging the expression and combining like terms:

$2x^2+6x-7=x^2+2x-2 \\ 2x^2+6x-7-x^2-2x+2=0 \\ x^2+4x-5=0$

Note that the coefficient of the squared term is 1, therefore, we can (try to) factor the expression on the left side using quick trinomial factoring:

Let's look for a pair of numbers whose product equals the free term in the expression, and whose sum equals the coefficient of the first-degree term, meaning two numbers $m,\hspace{2pt}n$ that satisfy:

$m\cdot n=-5\\ m+n=4\\$ From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for needs to yield a negative result. Therefore we can conclude that the two numbers have different signs, according to multiplication rules. Remember that the possible factors of 5 are 5 and 1. Fulfilling the second requirement mentioned, along with the fact that the numbers we're looking for have different signs will lead to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=5\\ n=-1\end{cases}$

Therefore we'll factor the expression on the left side of the equation to:

$x^2+4x-5=0 \\ \downarrow\\ (x+5)(x-1)=0$

Remember that the result of multiplication between expressions will yield 0 only if at least one of the multiplied expressions equals zero,

Therefore we obtain two simple equations and solve them by isolating the unknown term:

$x+5=0\\ \boxed{x=-5}$

or:

$x-1=0\\ \boxed{x=1}$

Let's summarize then the solution of the equation:

$2x^2+6x-7=x^2+2x-2\\ x^2+4x-5=0 \\ \downarrow\\ (x+5)(x-1)=0 \\ \downarrow\\ x+5=0\rightarrow\boxed{x=-5}\\ x-1=0\rightarrow\boxed{x=1}\\ \downarrow\\ \boxed{x=-5,1}$

Therefore the correct answer is answer A.

3

Final Answer

$x_1=-5,x_2=1$

Key Points to Remember

Essential concepts to master this topic

Rule: Move all terms to one side to get standard form
Technique: Factor $x^2+4x-5$ into $(x+5)(x-1)$
Check: Substitute x = -5: $2(-5)^2+6(-5)-7 = 50-30-7 = 13$ ✓

Common Mistakes

Avoid these frequent errors

Not combining like terms correctly when rearranging
Don't just move terms without changing signs = wrong equation! When moving $x^2+2x-2$ to the left side, you must subtract each term. Always change signs when moving terms across the equals sign.

Practice Quiz

Test your knowledge with interactive questions

$$ x^2-3x-18=0 $$

FAQ

Everything you need to know about this question

How do I know which two numbers to use for factoring?

+

Look for two numbers that multiply to give the constant term (-5) and add to give the middle coefficient (4). For $x^2+4x-5$ , you need 5 × (-1) = -5 and 5 + (-1) = 4.

What if the quadratic doesn't factor easily?

+

If you can't find two numbers that work, try the quadratic formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ . This works for any quadratic equation!

Why do I get two answers for x?

+

Quadratic equations usually have two solutions because a parabola can cross the x-axis at two points. When $(x+5)(x-1) = 0$ , either factor can equal zero.

How do I check if both solutions are correct?

+

Substitute each value back into the original equation. For x = -5: both sides equal 13. For x = 1: both sides equal 1. If both check out, you have the right answers!

Do I always need to get zero on one side?

+

Yes! For factoring to work, you need the equation in the form $ax^2 + bx + c = 0$ . This is called standard form and is essential for the factoring method.

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Solve 2x²+6x-7 = x²+2x-2: Finding the Value of X

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