Question

Solve the quadratic equation by completing the square.

$2x² - 20x + 48 = 0$
​

Answer 1

Question :

• Solve the quadratic equation by completing the square. 2x² - 20x + 48 = 0.

Solution :

We know that,

• A quadratic equation is in the form of ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0.

So,

→ 2x² - 20x + 48 = 0

→ x² - 10x + 24 = 0

→ x² - 4x - 6x + 24 = 0

→ x(x - 4) - 6x + 24 = 0

→ x(x - 5) - 6(x - 4) = 0

→ (x - 4)(x - 6) = 0

→ (x - 4) = 0 or (x - 6) = 0

→ x = 4 or (x - 6) = 0

→ x = 4 or x = 6 (Ans.)

Alternative method:

→ 2x² - 20x - 48 = 0

acx² + (ad + bc)x + bd = (ax + b)(cx + d);

→ 2(x - 6)(x - 4) = 0

if the product of the factor is 0, at least one factor should be 0;

→ (x - 6) = 0 or (x - 4) = 0

Solve the equation to find x,

→ x = 6 or x = 4 (Ans.)

Hence, the 6 and 4 are the roots of the given quadratic equation.

Learn more :

1. solve the quadratic equation√3x²+ 4x-7√3=0 by factorisation method

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2. The zeroes of the polynomial x2 + x - 2are

0 2,1

O 2-1

0 2,1

0 -21

brainly.in/question/31939737

Answer 2

Given : Quadratic Equation : 2x² - 20x + 48 = 0

Exigency to Solve : The Given Quadratic Equation .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad \dag\:\:\bigg\lgroup \bf{ \:Quadratic \:Equation\:: 2x^2 - 20x + 48 = 0 }\bigg\rgroup$

$\Large {\gray{\bf { \qquad \:Let's \: Solve \:The\:Quadratic \:Equation \::}}}$

$\qquad \dag\:\:\bf Quadratic \:Equation \::2x^2 - 20x + 48 = 0$

$\qquad \longmapsto \sf 2x^2 - 20x + 48 = 0 \\\\ \sf \star \:\:By \:taking\:2 \:as\:common \:in\:each\:term\:: \\\\\longmapsto \sf 2x^2 - 20x + 48 = 0\\\\ \longmapsto \sf x^2 - 10x + 24 = 0 \\\\\sf\star\:\: By \:Using\:Sum\:Product\:Pattern\::\\\\ \longmapsto \sf x^2 - 10x + 24 = 0\\\\ \longmapsto \sf x^2 - 4x - 6x + 24 = 0 \\\\ \sf \star\:\: By \:Finding\:out\:common\:Factors \:\:: \\\\ \longmapsto \sf x^2 - 4x - 6x + 24 = 0 \\\\ \longmapsto \sf x(x - 4) -6 ( x - 4) = 0\\\\ \star\:\:\sf Now,\:Rewrite\:in\:Factored\:term\:\;: \\\\\longmapsto \sf x(x - 4) -6 ( x - 4) = 0\\\\ \longmapsto \sf (x - 4) (x - 6) = 0$

$\qquad \longmapsto \frak{\underline{\purple{\:x = 4 \:\:\:or\;\:\: \:6 \:}} }\bigstar$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The \:Root\:of \:Quadratic \:Equation \:are\:\bf{6 \:\& \:4 .}}}}$

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M O R E ⠀T O⠀K N O W :

find zeros of the polynomial of

• 28x²+37x-21

https://brainly.in/question/37481095

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