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Question: What is the quadratic formula for \(2{{x}^{2}}-3x-5=0\) ?...

What is the quadratic formula for 2x2−3x−5=02{{x}^{2}}-3x-5=0 ?

Explanation

Solution

From the given question we have to find the solutions for the 2x2−3x−5=02{{x}^{2}}-3x-5=0 by using the quadratic formula. The quadratic formula helps us to solve any quadratic equation. First, we bring the equation to the form ax2+bx+c=0a{{x}^{2}}+bx+c=0, where a, b, and c are coefficients. Then, we plug these coefficients in the formula −b±b2−4ac2a\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}. By using this formula, we will get the solutions for the above quadratic equation. here we have to proceed further by comparing the equation with the general equation and then we have to apply the quadratic formula.

Complete step-by-step solution:
From the question given equation is
⇒2x2−3x−5=0\Rightarrow 2{{x}^{2}}-3x-5=0
Now we have to solve this equation by using quadratic formula as we all know that,
The quadratic formula helps us to solve any quadratic equation. First, we bring the equation to the general form which is ax2+bx+c=0a{{x}^{2}}+bx+c=0, where a, b, and c are coefficients. Then, we plug these coefficients in the formula
The formula is
⇒−b±b2−4ac2a\Rightarrow \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
Now the given quadratic equation is in the form of ax2+bx+c=0a{{x}^{2}}+bx+c=0 then by comparing the given equation with the general quadratic equation we will get the values of a, b and c.
By comparing we will get,
⇒a=2\Rightarrow a=2
⇒b=−3\Rightarrow b=-3
⇒c=−5\Rightarrow c=-5
Now we have to substitute the above values in their respective positions in the formula.
By substituting the above values in their respective positions in the formula we will get,
⇒−b±b2−4ac2a\Rightarrow \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
⇒x=−(−3)±(−3)2−4×2×−52×2\Rightarrow x=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\times 2\times -5}}{2\times 2}
⇒x=3±9+404\Rightarrow x=\dfrac{3\pm \sqrt{9+40}}{4}
⇒x=3±494\Rightarrow x=\dfrac{3\pm \sqrt{49}}{4}
⇒x=3±74\Rightarrow x=\dfrac{3\pm 7}{4}
⇒x=3+74orx=3−74\Rightarrow x=\dfrac{3+7}{4}\quad or\quad x=\dfrac{3-7}{4}
⇒x=104orx=−44\Rightarrow x=\dfrac{10}{4}\quad or\quad x=\dfrac{-4}{4}
⇒x=52orx=−1\Rightarrow x=\dfrac{5}{2}\quad or\quad x=-1
Therefore, the solutions of the above equation 2x2−3x−5=02{{x}^{2}}-3x-5=0 by using quadratic formula is x=52orx=−1 x=\dfrac{5}{2}\quad or\quad x=-1.

Note: Students should know the basic formulas of quadratic equations. In the formula there is ±\pm students should remember this if they only write plus, they will only get one root.
As in the above question they asked to do by using quadratic formula without this also by
factorization we can solve this equation. in the following way,
⇒2x2−3x−5=0 ⇒2x2−5x+2x−5=0 ⇒2x(x+1)−5(x+1)=0 ⇒(x+1)(2x−5)=0 ⇒x=52orx=−1 \begin{aligned} & \Rightarrow 2{{x}^{2}}-3x-5=0 \\\ & \Rightarrow 2{{x}^{2}}-5x+2x-5=0 \\\ & \Rightarrow 2x\left( x+1 \right)-5\left( x+1 \right)=0 \\\ & \Rightarrow \left( x+1 \right)\left( 2x-5 \right)=0 \\\ & \Rightarrow x=\dfrac{5}{2}\quad or\quad x=-1 \\\ \end{aligned}