
Solve: ${x^2} - 5x - 14 = 0$
Answer
559.5k+ views
Hint:
In this question first we will multiply the constant term and the coefficient of ${x^2}$ and then we will factorize the result of the multiplication and finally we will try to write the coefficient of $x$ in terms of the factors we got by factoring the result of the multiplication.
Complete step by step solution:
The given equation is ${x^2} - 5x - 14 = 0$. The constant value of the equation is $14$ and the coefficient of ${x^2}$ in the equation is $1$. Now, multiply these two values $14 \times 1 = 14$. Now factorize $14$ therefore it can be written as $7 \times 2$ . Now, if we observe that the coefficient of $x$ is $ - 5$ and it can be formed by $7$ and $2$ i.e. $ - 7 + 2 = - 5$. Now write the middle term of the equation ${x^2} - 5x - 14 = 0$ as ${x^2} - 7x + 2x - 14 = 0$. Therefore, we can write:
$
\Rightarrow {x^2} - 7x + 2x - 14 = 0 \\
\Rightarrow x\left( {x - 7} \right) + 2\left( {x - 7} \right) = 0 \\
$
Now, take $\left( {x - 7} \right)$ common from the above equation. Therefore, the above equation can be written as follows:
$ \Rightarrow \left( {x - 7} \right)\left( {x + 2} \right) = 0$
Now, we can write $\left( {x - 7} \right) = 0$ or $\left( {x + 2} \right) = 0$
Therefore, the values of $x$ are $x = 7$ or $x = - 2$.
Therefore, there are two solutions to the given equation.
Additional information:
Zeros of the polynomial are the values at which the value of the polynomial becomes zero.
Note:
In this question the important thing is the formation of the middle term of the equation with the help of the factors we have found because if we calculate it incorrect then the final answer will also be incorrect. So just be careful about these things. This question can also be solved by using the formula to find the roots of the quadratic equation. So please try to do this question by both the methods.
In this question first we will multiply the constant term and the coefficient of ${x^2}$ and then we will factorize the result of the multiplication and finally we will try to write the coefficient of $x$ in terms of the factors we got by factoring the result of the multiplication.
Complete step by step solution:
The given equation is ${x^2} - 5x - 14 = 0$. The constant value of the equation is $14$ and the coefficient of ${x^2}$ in the equation is $1$. Now, multiply these two values $14 \times 1 = 14$. Now factorize $14$ therefore it can be written as $7 \times 2$ . Now, if we observe that the coefficient of $x$ is $ - 5$ and it can be formed by $7$ and $2$ i.e. $ - 7 + 2 = - 5$. Now write the middle term of the equation ${x^2} - 5x - 14 = 0$ as ${x^2} - 7x + 2x - 14 = 0$. Therefore, we can write:
$
\Rightarrow {x^2} - 7x + 2x - 14 = 0 \\
\Rightarrow x\left( {x - 7} \right) + 2\left( {x - 7} \right) = 0 \\
$
Now, take $\left( {x - 7} \right)$ common from the above equation. Therefore, the above equation can be written as follows:
$ \Rightarrow \left( {x - 7} \right)\left( {x + 2} \right) = 0$
Now, we can write $\left( {x - 7} \right) = 0$ or $\left( {x + 2} \right) = 0$
Therefore, the values of $x$ are $x = 7$ or $x = - 2$.
Therefore, there are two solutions to the given equation.
Additional information:
Zeros of the polynomial are the values at which the value of the polynomial becomes zero.
Note:
In this question the important thing is the formation of the middle term of the equation with the help of the factors we have found because if we calculate it incorrect then the final answer will also be incorrect. So just be careful about these things. This question can also be solved by using the formula to find the roots of the quadratic equation. So please try to do this question by both the methods.
Recently Updated Pages
Two men on either side of the cliff 90m height observe class 10 maths CBSE

Cutting of the Chinese melon means A The business and class 10 social science CBSE

Show an aquatic food chain using the following organisms class 10 biology CBSE

How is gypsum formed class 10 chemistry CBSE

If the line 3x + 4y 24 0 intersects the xaxis at t-class-10-maths-CBSE

Sugar present in DNA is A Heptose B Hexone C Tetrose class 10 biology CBSE

Trending doubts
Why is there a time difference of about 5 hours between class 10 social science CBSE

What is the median of the first 10 natural numbers class 10 maths CBSE

Indias first jute mill was established in 1854 in A class 10 social science CBSE

Indias first jute mill was established in 1854 in A class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

