inimigurhkg

2023-02-25

How to solve $2{x}^{2}-7x+3=0$ by completing the square.

Cailyn Knight

Given: $2{x}^{2}-7x+3=0$
$⇒$$2\left({x}^{2}-\frac{7}{2}x\right)+3=0$
Making a perfect square by adding and subtracting $2{\left(\frac{7}{4}\right)}^{2}$
$2\left({x}^{2}-\frac{7}{2}x+{\left(\frac{7}{4}\right)}^{2}\right)-2{\left(\frac{7}{4}\right)}^{2}+3=0$
$⇒$$2{\left(x-\frac{7}{4}\right)}^{2}-\frac{49}{8}+3=0$
$⇒$$2{\left(x-\frac{7}{4}\right)}^{2}=\frac{25}{8}$
$⇒$${\left(x-\frac{7}{4}\right)}^{2}=\frac{25}{16}$
$⇒$$\left(x-\frac{7}{4}\right)=±\frac{5}{4}$
$⇒$$x=±\frac{5}{4}+\frac{7}{4}$
$⇒$$x=3or\frac{1}{2}$
Hence, the roots of $2{x}^{2}-7x+3=0$are $3$and $\frac{1}{2}.$

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