Question

solve the questions 2 x 2_ 7x _15 = 0 by method of completing the square ​

Answer 1

Answer:

$\bigstar{\bold{Zeros\:are\:\dfrac{3}{2} \:and\:-5}}$

Step-by-step explanation:

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

• p(x) = 2x²- 7x -15 = 0

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

• Zeros of the polynomial by completing the square method

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

→ First bring the constant term to the RHS side

   2x² - 7x = 15

→ Divide the whole equation by 2

  x² - 7x/2 = 15/2

→ Now take the middle term and divide it by 2

  7/2/2 = 7/4

→ Squaring it we get 7²/4² = 49/16

→ Add this on both LHS and RHS

  $x^{2} -\dfrac{7x}{2} +\dfrac{49}{16} =\dfrac{15}{2}+\dfrac{49}{16}$

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

$(x+\dfrac{7}{4}) ^{2} =\dfrac{15}{2} +\dfrac{49}{16}$

$(x+\dfrac{7}{4}) ^{2} =\dfrac{120}{16} +\dfrac{49}{16}$

$(x+\dfrac{7}{4}) ^{2} =\dfrac{169}{16}$

→ Taking root on both sides

$x + \dfrac{7}{4}=\pm \dfrac{13}{4}$

→ Case 1:

$x+\dfrac{7}{4} =\dfrac{13}{4}$

$x=\dfrac{13}{4}-\dfrac{7}{4}$

$x=\dfrac{6}{4} =\dfrac{3}{2}$

→ Case 2 :

$x+\dfrac{7}{4} =-\dfrac{13}{4}$

$x=-\dfrac{13}{4}-\dfrac{7}{4}$

$x=-\dfrac{20}{4} =\:-5$

$\boxed{\bold{Zeros\:are\:\dfrac{3}{2} \:and\:-5}}$

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

The zeros of a polynomial can be found out by

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