Answer
Verified
412.5k+ views
Hint: There are various ways which one might use to solve an equation, but first one needs to identify the highest power of the variable to be found in the equation. In this equation, we cannot directly find the highest power of x. Multiply the expressions to get a single line expression on the LHS. Proceed to the solution accordingly.
Complete step-by-step solution:
In the question we can see that there is no clarity on the highest power of the variable, x.
Hence, we use the BODMAS rule to open the brackets and multiply the contents with each other.
The LHS changes and the equation becomes,
$
(2x - 1)(x + 3) = 0 \\
\Rightarrow 2x \times x - 1 \times x + 3 \times 2x - 1 \times 3 = 0 \\
\Rightarrow 2{x^2} - x + 6x - 3 = 0 \\
\Rightarrow 2{x^2} + 5x - 3 = 0 \\
$
Thus, we see that the highest power of x is 2, which makes the equation a quadratic equation.
Now, we know that $x \times y = 0$ means that x and y both individually equate to zero.
So, from the question,
$(2x - 1) = 0,(x + 3) = 0$
$ \Rightarrow 2x = 1,x = - 3$
$ \Rightarrow x = \dfrac{1}{2},x = - 3$
Note: Mid-term factorization or completing square methods often become cumbersome and are time taking to solve. Whenever dealing with fractional or imaginary values, use of the quadratic formula to solve quadratic equations is recommended. The quadratic formula is $x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where a is the coefficient of ${x^2}$, b is the coefficient of x and c is the constant from the standard quadratic equation, $a{x^2} + bx + c = 0$.
Complete step-by-step solution:
In the question we can see that there is no clarity on the highest power of the variable, x.
Hence, we use the BODMAS rule to open the brackets and multiply the contents with each other.
The LHS changes and the equation becomes,
$
(2x - 1)(x + 3) = 0 \\
\Rightarrow 2x \times x - 1 \times x + 3 \times 2x - 1 \times 3 = 0 \\
\Rightarrow 2{x^2} - x + 6x - 3 = 0 \\
\Rightarrow 2{x^2} + 5x - 3 = 0 \\
$
Thus, we see that the highest power of x is 2, which makes the equation a quadratic equation.
Now, we know that $x \times y = 0$ means that x and y both individually equate to zero.
So, from the question,
$(2x - 1) = 0,(x + 3) = 0$
$ \Rightarrow 2x = 1,x = - 3$
$ \Rightarrow x = \dfrac{1}{2},x = - 3$
Note: Mid-term factorization or completing square methods often become cumbersome and are time taking to solve. Whenever dealing with fractional or imaginary values, use of the quadratic formula to solve quadratic equations is recommended. The quadratic formula is $x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ where a is the coefficient of ${x^2}$, b is the coefficient of x and c is the constant from the standard quadratic equation, $a{x^2} + bx + c = 0$.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE