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

Question

5x2+15x40=3x+x2 5x^2+15x-40=3x+x^2

Video Solution

Solution Steps

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 Solution

Let's solve the given equation:

5x2+15x40=3x+x2 5x^2+15x-40=3x+x^2

First, let's organize the equation by moving terms and combining like terms:

5x2+15x40=3x+x25x2+15x403xx2=04x2+12x40=0 5x^2+15x-40=3x+x^2 \\ 5x^2+15x-40-3x-x^2=0 \\ 4x^2+12x-40=0

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

4x2+12x40=0/:4x2+3x10=0 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,n m,\hspace{2pt}n that satisfy:

mn=10m+n=3 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, and now we'll remember that the possible factors of 10 are 5 and 2 or 10 and 1, fulfilling the second requirement mentioned, together 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:

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

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

x2+3x10=0(x+5)(x2)=0 x^2+3x-10=0 \\ \downarrow\\ (x+5)(x-2)=0

From here we'll remember that the product of expressions equals 0 only if at least one of the multiplied expressions equals zero,

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

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

or:

x2=0x=2 x-2=0\\ \boxed{x=2}

Let's summarize the solution of the equation:

5x2+15x40=3x+x24x2+12x40=0x2+3x10=0(x+5)(x2)=0x+5=0x=5x2=0x=2x=5,2 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.

Answer

x1=5,x2=2 x_1=-5,x_2=2