Question

solve the following equation by quadratic formula:
$\dfrac{{x + 3}}{{x + 2}} = \dfrac{{3x - 7}}{{2x - 3}}\,\,;\,\,x \ne - 2,\dfrac{3}{2}$

Answer 1

Hint: This is a simple question of the quadratic equation. We just have to find the roots of our quadratic equations. But first, we have to make a quadratic equation. Then we can solve our problem either by completing the square method or dharacharya formula or factorization method. One more important thing to remember is that whenever we take the root of any number, it comes with both positive and negative signs.
Formula used: $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$D = {b^2} - 4ac$

Complete step by step answer:
In the given question,
$\dfrac{{x + 3}}{{x + 2}} = \dfrac{{3x - 7}}{{2x - 3}}$
On cross-multiplication, we get
$\left( {x + 3} \right)\left( {2x - 3} \right) = \left( {3x - 7} \right)\left( {x + 2} \right)$
On multiplying, we get
$2{x^2} - 3x + 6x - 9 = 3{x^2} + 6x - 7x - 14$
$2{x^2} + 3x - 9 = 3{x^2} - x - 14$
On transposing L.H.S to R.H.S, we get
$3{x^2} - 2{x^2} - x - 3x - 14 + 9 = 0$
${x^2} - 4x - 5 = 0$
Now,
Let’s calculate D,
$D = {b^2} - 4ac$
On comparing $a{x^2} + bx + c = 0$ with the above equation
$a = 1$
$b = - 4\,\,,\,\,\,c = - 5$
$D = {\left( { - 4} \right)^2} - 4 \times 1 \times \left( { - 5} \right)$
$D = 16 + 20$
$D = 36$
Now,
Let’s calculate the value of x
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
On putting the values of a, b and D.
We get,
$x = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {36} }}{{2 \times 1}}$
Taking the square root of $36$
$x = \dfrac{{4 \pm 6}}{2}$
$x = \dfrac{4}{2} \pm \dfrac{6}{2}$
$x = 2 \pm 3$
So, here are two values of x one with a sign and one with a sign.
$x = 2 + 3\,,\,\,2 - 3$
$x = 5, - 1$
Hence, two values of x are $5$ and $ - 1$ .

Note:
While solving the questions in which some restrictions are given one need to be very careful because we can’t take those values as answers if they are coming out to be. There are multiple ways to solve a question of a quadratic equation like dharacharya formula, completing the square method, and factorization method.