Question

Solve 8x^2+10x +3 using completing the square method

Answer 1

Answer

x = -1/2 , x = -3/4

Given

• 8x² + 10x + 3 = 0

To Find

To solve the given equation using completing square method

Solution

$\rm 8x^2+10x+3=0\\\\\implies \rm \dfrac{8x^2}{8}+\dfrac{10x}{8}+\dfrac{3}{8}=0\ [Divide\ with\ 8]\\\\\implies \rm x^2+2.x.\dfrac{5}{8}=-\dfrac{3}{8}\\\\\implies \rm x^2+2.x.\dfrac{5}{8}+\bigg(\dfrac{5}{8}\bigg)^2=-\dfrac{3}{8}+\bigg(\dfrac{5}{8}\bigg)^2\\\\\rm Since,Adding\ \bigg(\dfrac{5}{8}\bigg)^2\ on\ both\ sides\\\\\implies \rm \bigg(x+\dfrac{5}{8}\bigg)^2=-\dfrac{3}{8}+\dfrac{25}{64}\\\\\implies \rm \bigg(x+\dfrac{5}{8}\bigg)^2=\dfrac{-24+25}{64}$

$\implies \rm \bigg(x+\dfrac{5}{8}\bigg)^2=\dfrac{1}{64}\\\\\implies \rm x+\dfrac{5}{8}=\pm \dfrac{1}{8}\\\\\implies \rm x=\dfrac{\pm1-5}{8}\\\\\implies \rm x=\dfrac{-1-5}{8}\ \&\ x=\dfrac{+1-5}{8}\\\\\implies \rm x=\dfrac{-6}{8}\ \&\ x=\dfrac{-4}{8}\\\\\implies \bf x=-\dfrac{3}{4}\ \&\ x=-\dfrac{1}{2}$

Answer 2

Answer:

$\bigstar{\bold{Zeros\:are\:-\dfrac{1}{2} \:and-\dfrac{3}{4} }}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• p(x) = 8x² + 10x + 3

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Zeros of the polynomial by completing the square method

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ First bring the constant term to the RHS

  8x² + 10x = -3

→ Divide the whole equation with 8

  $x^{2} +\dfrac{5}{4} \times x=\dfrac{-3}{8}$

→ Next take the middle term and divide it by 2

   5/4 /2 = 5/8

→ Squaring it we get 25/64

→ Add this on both LHS and RHS

$x^{2} +\dfrac{5}{4} x+\dfrac{25}{64} =\dfrac{-3}{8} +\dfrac{25}{64}$

→ The LHS is in the form (a² + 2ab + b²). Converting it to (a+b)²

   $(x+\dfrac{5}{8} )^{2} =\dfrac{1}{64}$

→ Taking root on both sides

   $x+\dfrac{5}{8} =\pm \dfrac{1}{8}$

→ Case 1 :

$x+\dfrac{5}{8} =\dfrac{1}{8}$

$x=\dfrac{1}{8} -\dfrac{5}{8}$

$x = \dfrac{-4}{8} =\dfrac{-1}{2}$

→ Case 2:

$x+\dfrac{5}{8} =-\dfrac{1}{8}$

$x = -\dfrac{1}{8} -\dfrac{5}{8}$

$x = -\dfrac{6}{8} = -\dfrac{3}{4}$

$\boxed{\bold{Zeros\:are\:-\dfrac{1}{2} \:and-\dfrac{3}{4} }}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The zeros of a polynomial can be found out by

• Factorization method
• Splitting the middle term
• Completing the square