Solve the Quadratic Equation: 5x²+15x-40=3x+x²

Q: Why do I need to rearrange the equation first?

You need standard form ($ ax^2+bx+c=0 $) to use factoring methods. When terms are on both sides, you can't identify the correct coefficients for factoring.

Q: How do I know which numbers multiply to give the constant term?

For $ x^2+3x-10=0 $, find two numbers that multiply to -10 and add to +3. Since the product is negative, the numbers have opposite signs: +5 and -2 work perfectly!

Q: What if I can't factor the quadratic?

If factoring doesn't work easily, you can always use the quadratic formula: $ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $. It works for any quadratic equation!

Q: Why do I get two answers?

Quadratic equations typically have two solutions because a parabola can cross the x-axis at two points. Both answers are correct unless the problem asks for a specific one!

Q: Do I need to simplify by dividing by 4 like in the solution?

It's not required, but it makes factoring much easier! Working with $ x^2+3x-10=0 $ instead of $ 4x^2+12x-40=0 $ saves time and reduces errors.

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

00:22 Collect terms

00:36 Take out 4 from the parentheses

00:43 Pay attention to the trinomial coefficients

00:51 We want to find 2 numbers

00:55 Whose sum equals B and their product equals C

01:05 These are the appropriate numbers

01:16 Therefore these are the numbers we'll put in parentheses

01:27 Find the solutions that zero out each factor

01:33 Isolate X, this is the first solution

01:36 Isolate X, this is the second solution

01:40 And this is the solution to the problem

Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

Solve the following problem:

_{$5x^2+15x-40=3x+x^2$}

2

Step-by-step solution

Solve the given equation:

$5x^2+15x-40=3x+x^2$

Organize the equation by moving and combining like terms:

$5x^2+15x-40=3x+x^2 \\ 5x^2+15x-40-3x-x^2=0 \\ 4x^2+12x-40=0$

Note that all coefficients as well as the free term are multiples of 4, hence we'll divide both sides of the equation by 4:

$4x^2+12x-40=0\hspace{6pt}\text{/}:4 \\ x^2+3x-10=0$

Now we notice 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 the desired values:

$m\cdot n=-10\\ m+n=3\\$ From the first requirement mentioned, that is - from the multiplication, we notice that the product of the numbers we're looking for must be negative. Therefore we can conclude that the two numbers have different signs, according to multiplication rules. Remember that the possible factors of 10 are 5 and 2 or 10 and 1, fulfilling the second requirement mentioned. This together with the fact that the numbers we're looking for have different signs leads us to the conclusion that the only possibility for the two numbers we're looking for is:

$\begin{cases} m=5 \\ n=-2 \end{cases}$

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

$x^2+3x-10=0 \\ \downarrow\\ (x+5)(x-2)=0$

Remember that the product of expressions equals 0 only if at least one of the multiplied expressions equals zero,

Therefore we'll obtain two simple equations and solve them by isolating the variable in each:

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

or:

$x-2=0\\ \boxed{x=2}$

Let's summarize the solution of the equation:

$5x^2+15x-40=3x+x^2 \\ 4x^2+12x-40=0\\ x^2+3x-10=0 \\ \downarrow\\ (x+5)(x-2)=0 \\ \downarrow\\ x+5=0\rightarrow\boxed{x=-5}\\ x-2=0\rightarrow\boxed{x=2}\\ \downarrow\\ \boxed{x=-5,2}$

Therefore the correct answer is answer A.

3

Final Answer

$x_1=-5,x_2=2$

Key Points to Remember

Essential concepts to master this topic

Rearrangement: Move all terms to one side to get standard form
Technique: Factor $x^2+3x-10$ as $(x+5)(x-2)$
Check: Substitute $x=-5$ : $5(-5)^2+15(-5)-40 = 3(-5)+(-5)^2$ gives $35=35$ ✓

Common Mistakes

Avoid these frequent errors

Forgetting to move all terms to one side before factoring
Don't try to factor $5x^2+15x-40=3x+x^2$ directly = impossible to solve! The equation isn't in standard form. Always rearrange to get $ax^2+bx+c=0$ first, then factor.

Practice Quiz

Test your knowledge with interactive questions

$$ x^2+6x+9=0 $$

What is the value of X?

FAQ

Everything you need to know about this question

Why do I need to rearrange the equation first?

+

You need standard form ( $ax^2+bx+c=0$ ) to use factoring methods. When terms are on both sides, you can't identify the correct coefficients for factoring.

How do I know which numbers multiply to give the constant term?

+

For $x^2+3x-10=0$ , find two numbers that multiply to -10 and add to +3. Since the product is negative, the numbers have opposite signs: +5 and -2 work perfectly!

What if I can't factor the quadratic?

+

If factoring doesn't work easily, you can always use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ . It works for any quadratic equation!

Why do I get two answers?

+

Quadratic equations typically have two solutions because a parabola can cross the x-axis at two points. Both answers are correct unless the problem asks for a specific one!

Do I need to simplify by dividing by 4 like in the solution?

+

It's not required, but it makes factoring much easier! Working with $x^2+3x-10=0$ instead of $4x^2+12x-40=0$ saves time and reduces errors.

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Solve the Quadratic Equation: 5x²+15x-40=3x+x²

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