Solveeit Logo
Login

Question

Question: How do you factor \({{x}^{2}}-15x+50\)?...

How do you factor x2−15x+50{{x}^{2}}-15x+50?

Explanation

Solution

Here in this question, a quadratic equation is given in the form of ax2+bx+c=0a{{x}^{2}}+bx+c=0. We need to solve the given equation in order to find the roots or zeros. To solve a quadratic equation, we have to perform middle term splitting of -15x. This is the easiest approach to find the roots of the quadratic equation.

Complete step by step answer:
Now, let’s solve the question.
As we know that the equation having the highest degree of 2, and which is in the form ax2+bx+c=0a{{x}^{2}}+bx+c=0 is called a quadratic equation where a and b are coefficients and c is the constant term. Common method of finding roots of quadratic equations is splitting the middle term. In middle term splitting, first we have to find the product of the first and last term. And then find the factors of the product in such a way that the sum or difference will result in the middle term. Let’s see how it actually happens:
Step 1: write the quadratic equation.
⇒x2−15x+50\Rightarrow {{x}^{2}}-15x+50
Step 2: find the product of a and c where a is the coefficient of x2{{x}^{2}} and c is the constant.
So the product is 50×1=5050\times 1=50.
Step 3: as the factors of 50 are 10 and 5, so if we add 10 + 5 which is equal to 15. It gives the middle term. Now split the middle term:
⇒x2−10x−5x+50\Rightarrow {{x}^{2}}-10x-5x+50
Step 4: Now, group those terms from which common terms can be taken out.
⇒x(x−10)−5(x−10)\Rightarrow x(x-10)-5(x-10)
Step 5: Now, take out the common terms.
⇒(x−5)(x−10)\Rightarrow (x-5)(x-10)
Step 6: if each factor is equated to 0, then we can easily get the 2 roots of the equation.
For (x – 5) = 0
∴x=5\therefore x=5
For (x – 10) = 0
∴x=10\therefore x=10 The roots of the equations are: x = 5, 10

Note: Though middle term splitting is the easiest method of all, but still you can go for an alternate method that is by applying quadratic formula.
The quadratic formula is:
⇒x=−b±b2−4ac2a\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
⇒x2−15x+50\Rightarrow {{x}^{2}}-15x+50
In this equation, a = 1, b = -15 and c = 50. Now, put all the values in the quadratic formula. We will get:
⇒x=−(−15)±(−15)2−4(1)(50)2(1)\Rightarrow x=\dfrac{-\left( -15 \right)\pm \sqrt{{{\left( -15 \right)}^{2}}-4\left( 1 \right)\left( 50 \right)}}{2\left( 1 \right)}
Solve further:
⇒x=15±225−2002\Rightarrow x=\dfrac{15\pm \sqrt{225-200}}{2}
On Solving the under root, we get:
⇒x=15±52\Rightarrow x=\dfrac{15\pm 5}{2}
On finding both the values for x:
⇒x=15+52=202\Rightarrow x=\dfrac{15+5}{2}=\dfrac{20}{2}
∴\therefore x = 10
⇒x=15−52=102\Rightarrow x=\dfrac{15-5}{2}=\dfrac{10}{2}
∴\therefore x = 5
So, x = 5, 10
We got the same values by both methods. You should also know the square and square roots of at least 1 to 20.