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Question: How do you solve \(2{{x}^{2}}-13x-7=0\)?...

How do you solve 2x2−13x−7=02{{x}^{2}}-13x-7=0?

Explanation

Solution

As the given is the quadratic equation so we will solve it by using the quadratic formula. We know that for an equation of the form ax2+bx+c=0a{{x}^{2}}+bx+c=0 the quadratic formula is given as x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} . By substituting the values and solving further we will get the desired answer.

Complete step-by-step solution:
We have been given an equation 2x2−13x−7=02{{x}^{2}}-13x-7=0.
We have to solve the given equation.
We know that quadratic formula for the general equation ax2+bx+c=0a{{x}^{2}}+bx+c=0 is given as x=−b±b2−4ac2ax=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.
Now, by comparing the given equation with general equation we get the values
a=2,b=−13,c=−7a=2,b=-13,c=-7
Now, substituting the values in the formula we will get
⇒x=−(−13)±(−13)2−4×2×(−7)2×2\Rightarrow x=\dfrac{-\left( -13 \right)\pm \sqrt{{{\left( -13 \right)}^{2}}-4\times 2\times \left( -7 \right)}}{2\times 2}
Now, simplifying the above obtained equation we will get
⇒x=13±169+564 ⇒x=13±2254 \begin{aligned} & \Rightarrow x=\dfrac{13\pm \sqrt{169+56}}{4} \\\ & \Rightarrow x=\dfrac{13\pm \sqrt{225}}{4} \\\ \end{aligned}
Now, simplifying the above obtained equation we will get
⇒x=13±154\Rightarrow x=\dfrac{13\pm 15}{4}
Now, considering both signs one by one we will get
⇒x=13+154,x=13−154\Rightarrow x=\dfrac{13+15}{4},x=\dfrac{13-15}{4}
Now, simplifying the above obtained equation we will get
⇒x=284,x=−24 ⇒x=7,x=−12 \begin{aligned} & \Rightarrow x=\dfrac{28}{4},x=\dfrac{-2}{4} \\\ & \Rightarrow x=7,x=\dfrac{-1}{2} \\\ \end{aligned}
Hence we get the solution of the given quadratic equation as x=7,−12x=7,\dfrac{-1}{2}.

Note: Alternatively we can also solve the given equation by splitting the middle term and find the factors. Then we can equate each factor to zero and solve the equation for x.
We have 2x2−13x−7=02{{x}^{2}}-13x-7=0.
Now, splitting the middle term of the given equation we will get
⇒2x2−(14x−x)−7=0\Rightarrow 2{{x}^{2}}-\left( 14x-x \right)-7=0
Now, simplifying the above obtained equation we will get
⇒2x2−14x+x−7=0\Rightarrow 2{{x}^{2}}-14x+x-7=0
Now, taking common terms out we will get
⇒2x(x−7)+1(x−7)=0\Rightarrow 2x\left( x-7 \right)+1\left( x-7 \right)=0
Again taking common factor out we will get
⇒(x−7)(2x+1)=0\Rightarrow \left( x-7 \right)\left( 2x+1 \right)=0
Now, equating each factor to zero we will get
⇒x−7=0 and 2x+1=0 ⇒x=7 and x=−12 \begin{aligned} & \Rightarrow x-7=0\text{ and }2x+1=0 \\\ & \Rightarrow x=7\text{ and }x=\dfrac{-1}{2} \\\ \end{aligned}
Hence above is the required solution.