Solveeit Logo
Login

Question

Question: How do you find the quadratic using the quadratic formula given \(4{{x}^{2}}+11x=3x-10\) over the se...

How do you find the quadratic using the quadratic formula given 4x2+11x=3x−104{{x}^{2}}+11x=3x-10 over the set of complex numbers?

Explanation

Solution

There are various methods to solve the quadratic equation like factorisation by splitting the middle term, forming the perfect square and Quadratic formula. Here in the question we are asked to use quadratic formula directly, so first we simplify the given quadratic equation and then compare the given quadratic equation with general form of quadratic equation as ax2+bx+c=0a{x}^{2}+b{x}+c=0 , now we use the quadratic formula as −b±b2−4ac2a\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.

Complete step-by-step solution:
Now consider the given equation 4x2+11x=3x−104{{x}^{2}}+11x=3x-10
Now we can see that the given equation is a quadratic equation in x.
We want to first bring the quadratic equation in general form ax2+bx+c=0a{{x}^{2}}+bx+c=0
Now consider 4x2+11x=3x−104{{x}^{2}}+11x=3x-10 .
On transposing all the terms of RHS in LHS we get,
⇒4x2+11x−3x+10=0 ⇒4x2+8x+10=0 \begin{aligned} & \Rightarrow 4{{x}^{2}}+11x-3x+10=0 \\\ & \Rightarrow 4{{x}^{2}}+8x+10=0 \\\ \end{aligned}
Hence we have the given equation in general form ax2+bx+c=0a{{x}^{2}}+bx+c=0 where a = 4, b = 8 and
c = 10.
Now we know that the solution to the equation ax2+bx+c=0a{{x}^{2}}+bx+c=0 is given by the formula −b±b2−4ac2a\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} . Hence substituting the values of a, b and c in the formula we get,
⇒x=−8±82−4(4)(10)2(4) ⇒x=−8±64−1608 ⇒x=−8±i968 \begin{aligned} & \Rightarrow x=\dfrac{-8\pm \sqrt{{{8}^{2}}-4\left( 4 \right)\left( 10 \right)}}{2\left( 4 \right)} \\\ & \Rightarrow x=\dfrac{-8\pm \sqrt{64-160}}{8} \\\ & \Rightarrow x=\dfrac{-8\pm i\sqrt{96}}{8} \\\ \end{aligned}
Hence we get the solution of the equation x=−8+i968x=\dfrac{-8+i\sqrt{96}}{8} and x=−8−i968x=\dfrac{-8-i\sqrt{96}}{8}.

Note: Now note that we have the discriminant D which is defined as D=b2−4acD={{b}^{2}}-4ac . When we have D < 0 then the roots of the equation are complex roots and hence we get both the roots as complex roots. Also note that if a + ib is the solution of the equation then its conjugate defined as a – ib is also always the solution of the equation. This is in fact true for all polynomials in one variable.