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How do you solve \[{{x}^{2}}-3x-4=0\] by quadratic formula?

seo-qna
Last updated date: 03rd Mar 2024
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IVSAT 2024
Answer
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Hint: We are given \[{{x}^{2}}-3x-4=0\] and to solve this we learn about the type of equation we are given then learn a number of solutions to the equation. We will learn how to factor the quadratic equation and we will use the middle term split to factor the term and we will simplify by taking common terms out. We will also use the zero product rule to get our answer. To be sure about your answer we can also check by putting the acquired value of the solution in the given equation and check whether they are the same or not.

Complete step-by-step solution:
We are given \[{{x}^{2}}-3x-4=0\] and we are asked to solve the given problem. First, we observe that it has a maximum power of ‘2’ so it is a quadratic equation. Now we should know that the quadratic equation has 2 solutions or we say an equation of power ‘n’ will have an ‘n’ solution. Now as it is a quadratic equation we will change it into standard form \[a{{x}^{2}}+bx+c=0.\] As we look closely our problem is already in standard form.
\[{{x}^{2}}-3x-4=0\]
Now, we have to solve the equation \[{{x}^{2}}-3x-4=0.\]
To solve this equation we first take the greatest common factor possibly available to the terms. As we can see that in \[{{x}^{2}}-3x-4=0,\] 1, – 3 and – 4 have nothing in common. So, the equation remains the same.
\[{{x}^{2}}-3x-4=0\]
Now, we are asked to solve our problem using the quadratic formula. For any quadratic equation, \[a{{x}^{2}}+bx+c=0\] the quadratic formula is given as
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
For our equation, \[{{x}^{2}}-3x-4=0,\] we have a = 1, b = – 3 and c = – 4. So, using these values in the above formula, we will get,
\[\Rightarrow x=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\times 1\times -4}}{2\times 1}\]
On simplifying, we get,
\[\Rightarrow x=\dfrac{3\pm \sqrt{25}}{2}\]
So, we get,
\[\Rightarrow x=\dfrac{3\pm 5}{2}\]
Simplifying further, we get,
\[\Rightarrow x=\dfrac{3+5}{2}=\dfrac{8}{2};x=\dfrac{3-5}{2}=\dfrac{-2}{2}\]
So, we get,
\[\Rightarrow x=4;x=-1\]
Hence, the solutions are x = 4 and x = – 1.

Note: We can also solve this using another method which is called by factoring. We factor using the middle term split. For the quadratic equation, \[a{{x}^{2}}+bx+c=0\] we will find such a pair of a term whose products is the same as \[a\times c\] and whose sum of difference will be equal to b. Now in \[{{x}^{2}}-3x-4=0\] we have a = 1, b = – 3 and c = – 4. So, we get,
\[a\times x=1\times -4=-4\]
We can see that \[1\times -4=-4\] and their sum is – 4 + 1 = – 3. So, we use this to split.
\[{{x}^{2}}-3x-4=0\]
\[\Rightarrow {{x}^{2}}+\left( -4+1 \right)x-4=0\]
Opening the brackets, we get,
\[\Rightarrow {{x}^{2}}-4x+x-4=0\]
Taking common in the first two and last two terms and we get,
\[\Rightarrow x\left( x-4 \right)+1\left( x-4 \right)=0\]
As x – 4 is common, so simplifying further, we get,
\[\Rightarrow \left( x+1 \right)\left( x-4 \right)=0\]
Using zero product rule which says if two terms product is zero that either one of them is zero. So, either
\[\Rightarrow x+1=0;x-4=0\]
So we get, x = – 1 and x = 4.
Hence, the solutions are x = – 1 and x = 4.



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