Question : Which of the following equations has 7 as a root?
Option 1: $3x^2-6x+2=0$
Option 2: $x^2-9x+14=0$
Option 3: $x^2-7x+10=0$
Option 4: $x^2+3x-12=0$
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Correct Answer: $x^2-9x+14=0$
Solution :
Given:
Option 1:
$3x^2-6x+2=0$
Using the quadratic formula, we get,
$⇒x=\frac{6\pm\sqrt{(-6)^2-4.3.2}}{2.3}$
So, $x=\frac{\sqrt3+3}{3}$ or $x=\frac{-\sqrt{3}+3}{3}$
So, the value of $x\neq7$.
Option 2:
$x^2-9x+14=0$
$⇒x^2-7x-2x+14=0$
$⇒(x-7)(x-2)=0$
$\therefore x = 7$ or $x = 2$
Here, the value of $x=7$.
Hence, the correct answer is $x^2-9x+14=0$.
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