Question

How do you solve $2{x^2} - 9x - 5 = 0$ by factoring?

Answer 1

Hint: In this question, we have been given an equation and we have been asked to factorize it. To solve a quadratic equation by using factorization method, follow the following steps:
First, we will write the given equation in general form $a{x^2} + bx + c = 0$ .
Split the middle term and factorize the given equation.
After factoring, write the given quadratic equation as a product of two linear factors.
And then the final step is to equate each linear factor to zero and solve for x. These values of x will give the roots of the equation.

Complete step-by-step solution:
The quadratic equation is $2{x^2} - 9x - 5 = 0$
Now, we will split the middle term by taking two factors of $ - 10\left( {2 \times - 5} \right)$ such that when they are added or subtracted, they give us $ - 9$ . Such factors are $1$ and $ - 10$ .
$2{x^2} - 10x + x - 5 = 0$
Take $2x$ common in the first term and second term,
$2x(x - 5) + 1(x - 5) = 0$
Take common term $(x - 5)$
$(2x + 1)(x - 5) = 0$
Now, we will keep each factor equal to 0,
$2x + 1 = 0$ or $x - 5 = 0$
Finding the values of $x$ by shifting the constant terms to the other side.
$2x = - 1$ or $x = 5$
$x = - \dfrac{1}{2}$ or $x = 5$

Therefore, the solution is $x = - \dfrac{1}{2}$ or $x = 5$

Note: Some equations that are not quadratic can be solved by reducing to a quadratic equation by suitable substitutions. Example ${x^4} - 13{x^2} + 42$ The equation is not quadratic, so reducing the quadratic equation. ${({x^2})^2} - 13{x^2} + 42 = 0$. Then let,
${x^2} = a$ ,${a^2} - 13a + 42$.
Splitting the middle term, subtracting the two values are seven and six,
${a^2} - 7a - 6a + 42 = 0$
Now, take $a$ common in first and second term and $ - 6$ in third and fourth term
$a(a - 7) - 6(a - 7) = 0$
Take $(a - 7)$ common in first and second term
$(a - 7)(a - 6) = 0$
Splitting $x$ values from first term and, and then putting each factor equal to zero,
$a - 7 = 0$ or $a - 6 = 0$
Finding the values of a by shifting the constant to the other side.
$a = 7$ or $a = 6$
Now, we get
 $a = 7$ or $a = 6$
Since $a = {x^2}$ ,
If $x = 7$ then, ${x^2} = 7$ ,$x = \pm \sqrt 7 $
If $x = 6$ then, ${x^2} = 6$, $x = \pm \sqrt 6 $ .
Therefore, the solution is - $x = \pm \sqrt 7 $or $x = \pm \sqrt 6 $ .