Solve the quadratic equation: 2x^2

Q. Solve the quadratic equation: 2 X^2 – 7 X + 3 = 0

Ans: Given equation is:

2 X ^2 – 7 X + 3 = 0

∴ 2 X ^2 – 6 X – X + 3 = 0

∴ 2 X (X – 3) – 1 (X – 3) = 0

∴ (X – 3) (2 X – 1) = 0

∴ X = 3 and X = $\frac{1}{2}$

Therefore values of X are 3 and $\frac{1}{2}$

Check: Let’s check our answers:

at X = 3 in LHS, equation will become: 2 (3)² – 7 (3) + 3

= 18 – 21 + 3 = 0

at X = 1/2 in LHS, equation will become: 2 ( $\frac{1}{2}$ )² – 7 ( $\frac{1}{2}$ ) + 3

= 2 $\frac{1}{4}$ – 7 $\frac{1}{2}$ + 3

= $\frac{1}{2}$ – $\frac{7}{2}$ + 3 = $\frac{7}{2}$ – $\frac{7}{2}$ = 0

Since values of LHS matches with given value on RHS, our both answers are correct.

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