KnowledgeBoat Logo
|

Mathematics

Solve the following equation using the formula :

115x2+53=23x\dfrac{1}{15}x^2 + \dfrac{5}{3} = \dfrac{2}{3}x

Quadratic Equations

1 Like

Answer

Given,

⇒115x2+53=23x⇒x2+2515=23x⇒3(x2+25)=30x⇒3x2+75=30x⇒3x2−30x+75=0⇒3(x2−10x+25)=0⇒x2−10x+25=0.\Rightarrow \dfrac{1}{15}x^2 + \dfrac{5}{3} = \dfrac{2}{3}x \\[1em] \Rightarrow \dfrac{x^2 + 25}{15} = \dfrac{2}{3}x \\[1em] \Rightarrow 3(x^2 + 25) = 30x \\[1em] \Rightarrow 3x^2 + 75 = 30x \\[1em] \Rightarrow 3x^2 - 30x + 75 = 0 \\[1em] \Rightarrow 3(x^2 - 10x + 25) = 0 \\[1em] \Rightarrow x^2 - 10x + 25 = 0.

Comparing x2 - 10x + 25 = 0 with ax2 + bx + c = 0 we get,

a = 1, b = -10, c = 25.

We know that,

x = −b±b2−4ac2a\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting values we get,

⇒x=−(−10)±(−10)2−4(1)(25)2(1)=10±100−1002=10±02=10+02 or 10−02=5,5.\Rightarrow x = \dfrac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(25)}}{2(1)} \\[1em] = \dfrac{10 \pm \sqrt{100 - 100}}{2} \\[1em] = \dfrac{10 \pm \sqrt{0}}{2} \\[1em] = \dfrac{10 + 0}{2} \text{ or } \dfrac{10 - 0}{2} \\[1em] = 5, 5.

Hence, x = 5, 5.

Answered By

2 Likes

Related Questions