Solveeit Logo
Login

Question

Question: What are the solutions of \(4{{x}^{2}}-7x=3x+24\) ?...

What are the solutions of 4x2−7x=3x+244{{x}^{2}}-7x=3x+24 ?

Explanation

Solution

From the question we have been asked to find the solutions of a quadratic equation. Here, given a quadratic equation expression, we have to simplify the expression and make it into a standard form of quadratic equation. If the quadratic equation is in the form of ax2+bx+c=0a{{x}^{2}}+bx+c=0, then we know that the roots of this quadratic equation are given by
⇒x=−b±b2−4ac2a\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}

Complete step-by-step solution:
So, from the question we have that,
⇒4x2−7x=3x+24\Rightarrow 4{{x}^{2}}-7x=3x+24
Now we will group all the like terms like x2{{x}^{2}} terms x terms and constants as shown below:
⇒4x2−7x−3x−24=0\Rightarrow 4{{x}^{2}}-7x-3x-24=0
Now we will simplify the above equation, so the equation will be as given below,
⇒4x2−10x−24=0\Rightarrow 4{{x}^{2}}-10x-24=0
Now the above equation is in the standard form of a quadratic equation, which is ax2+bx+c=0a{{x}^{2}}+bx+c=0
Here comparing the equation 4x2−10x−24=04{{x}^{2}}-10x-24=0 with the standard form ax2+bx+c=0a{{x}^{2}}+bx+c=0 and compare the coefficients a, b and c.
⇒a=4,b=−10,c=−24\Rightarrow a=4,b=-10,c=-24
Now applying the formula to find the value of the roots of x, as given below
⇒x=−b±b2−4ac2a\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
Substituting the values of a band c in the above formula
⇒x=−(−10)±(−10)2−4×4×−242×4\Rightarrow x=\dfrac{-\left( -10 \right)\pm \sqrt{{{\left( -10 \right)}^{2}}-4\times 4\times -24}}{2\times 4}
Simplifying the above expression, as given below
⇒x=10±100+3848\Rightarrow x=\dfrac{10\pm \sqrt{100+384}}{8}
⇒x=10±228\Rightarrow x=\dfrac{10\pm 22}{8}
Now considering the two cases, with plus and minus, as shown
⇒x=10+228;x=10−228\Rightarrow x=\dfrac{10+22}{8};x=\dfrac{10-22}{8}
⇒x=328;x=−128\Rightarrow x=\dfrac{32}{8};x=\dfrac{-12}{8}
⇒x=4;x=−32\Rightarrow x=4;x=\dfrac{-3}{2}
**Hence the value of the roots or the solutions are equal to
⇒x=4;x=−32\Rightarrow x=4;x=\dfrac{-3}{2} **

Note: Please note that this problem can also be done either by the method of completing the square or just factoring and solving the quadratic equation to solve ax2+bx+c=0a{{x}^{2}}+bx+c=0 by completing the square transform the equation so that the constant term c is alone or the right side. But here we are adding and subtracting some terms in order to factor.