Question

Solve $7{x^2} - 19x - 6 = 0$ .

Answer 1

Hint: In this question, we are given a quadratic equation which we have to solve for $x$ .
So, we can do this either by splitting the middle term or by the discriminant method.
Consider, a quadratic equation, $a{x^2} + bx + c = 0$ , in splitting middle term, we have to find a pair of numbers, say $p,q$ , whose product is equal to the product of $a,c$ i.e., $p \times q = a \times c$ and whose sum is equal to $b$ , i.e., $p + q = b$ .

Complete step-by-step answer:
Given equation: $7{x^2} - 19x - 6$ .
To solve the given equation for $x$ .
Now, first, we have to find the pair $p,q$ whose product is equal to $7 \times \left( { - 6} \right) = - 42$ and whose sum is equal to $ - 19$.
Since, we know, $7 \times 6 = 42$ and prime factors of $42$ are $2,3,7$ . So, using these factors, we get, $7 \times 3 = 21$ and $2 - 21 = - 19$ . Hence, the pair is $2,21$ .
Now, we can rewrite the equation as $7{x^2} + 2x - 21x - 6 = 0$ . Now, take $x$ common from the first two terms and $ - 3$ from the last two terms, then, the equation becomes, $x\left( {7x + 2} \right) - 3\left( {7x + 2} \right) = 0$ .
Now, take $\left( {7x + 2} \right)$ common from the whole equation, then we get, $\left( {7x + 2} \right)\left( {x - 3} \right) = 0$ .
Now, either $\left( {7x + 2} \right) = 0$ or $\left( {x - 3} \right) = 0$ .
So, if $\left( {7x + 2} \right) = 0$ , then we have $x = \dfrac{{ - 2}}{7}$ , and if $\left( {x - 3} \right) = 0$ , then we have $x = 3$ .
Hence, $x$ is either equal to $\dfrac{{ - 2}}{7}$ or equal to $3$ .

Note: Another way to solve this equation is the discriminant method, in which, for a quadratic equation $a{x^2} + bx + c = 0$ , $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ . So, for given quadratic equation, $7{x^2} - 19x - 6$ , we have $a = 7,b = - 19,c = - 6$ , putting values in $x$ , we get, $x = \dfrac{{19 \pm \sqrt {{{\left( { - 19} \right)}^2} - 4\left( 7 \right)\left( { - 6} \right)} }}{{2\left( 7 \right)}}$ , on solving, we get the same answer as $x = \dfrac{{ - 2}}{7}{\text{ or 3}}$ .
This method is only valid for a quadratic equation.
One must remember that, if $p \times q = 0$ , then, either $p = 0$ or $q = 0$ .