
Solve the following:
\[(x - 3)(2x + 5) = 0\]
Answer
500.4k+ views
Hint: Find the product of the two binomial terms and form the expression in the form of quadratic expression and frame the simplified expression. Split the middle term and make a pair of terms and find the value for “x”.
Complete step-by-step answer:
Take the given expression: \[(x - 3)(2x + 5) = 0\]
Find the product of the two binomials for the above expression –
\[x(2x + 5) - 3(2x + 5) = 0\]
Multiply the term outside the bracket with the term inside the bracket.
\[2{x^2} + 5x - 6x - 15 = 0\]
The above expression can be re-written as –
\[2{x^2} - 6x + 5x - 15 = 0\]
Make the pair of first two and the last two terms.
\[\underline {2{x^2} - 6x} + \underline {5x - 15} = 0\]
Take the common multiple common from the paired terms in the above expression –
$2x(x - 3) + 5(x - 3) = 0$
Take the common multiple common from the above expression –
$(x - 3)(2x + 5) = 0$
There are two cases –
$x - 3 = 0$
Make the required term the subject, when you move any term from one side to the opposite side then the sign of the terms also changes. Positive term changes to negative and vice-versa.
$x = 3$ …. (A)
Now, second case
$
2x + 5 = 0 \\
2x = - 5 \;
$
Term multiplicative on one side if moved to the opposite side then it goes to the denominator.
$x = \dfrac{{ - 5}}{2}$ …. (B)
Hence, the required solution is $x = \dfrac{{ - 5}}{2}$or $x = 3$
So, the correct answer is “$x = \dfrac{{ - 5}}{2}$or $x = 3$”.
Note: A quadratic equation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real.
Complete step-by-step answer:
Take the given expression: \[(x - 3)(2x + 5) = 0\]
Find the product of the two binomials for the above expression –
\[x(2x + 5) - 3(2x + 5) = 0\]
Multiply the term outside the bracket with the term inside the bracket.
\[2{x^2} + 5x - 6x - 15 = 0\]
The above expression can be re-written as –
\[2{x^2} - 6x + 5x - 15 = 0\]
Make the pair of first two and the last two terms.
\[\underline {2{x^2} - 6x} + \underline {5x - 15} = 0\]
Take the common multiple common from the paired terms in the above expression –
$2x(x - 3) + 5(x - 3) = 0$
Take the common multiple common from the above expression –
$(x - 3)(2x + 5) = 0$
There are two cases –
$x - 3 = 0$
Make the required term the subject, when you move any term from one side to the opposite side then the sign of the terms also changes. Positive term changes to negative and vice-versa.
$x = 3$ …. (A)
Now, second case
$
2x + 5 = 0 \\
2x = - 5 \;
$
Term multiplicative on one side if moved to the opposite side then it goes to the denominator.
$x = \dfrac{{ - 5}}{2}$ …. (B)
Hence, the required solution is $x = \dfrac{{ - 5}}{2}$or $x = 3$
So, the correct answer is “$x = \dfrac{{ - 5}}{2}$or $x = 3$”.
Note: A quadratic equation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real.
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