Question:
Let $x = 2$ be a root of $y = 4x^2 - 14x + q = 0$. Then $y$ is equal to
Let $x = 2$ be a root of $y = 4x^2 - 14x + q = 0$. Then $y$ is equal to
Updated On: May 6, 2024
- (x - 2) (4x - 6)
- (x - 2) (4x + 6)
- (x - 2) (-4x - 6)
- (x - 2) (-4x + 6)
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The Correct Option is A
Solution and Explanation
We have
$y=4 x^{2}-14 x+q=0$
Since, $x=2$ is the root
$\therefore 4(2)^{2}-14(2)+q=0$
$\Rightarrow 16-28+q=0$
$\Rightarrow q=12$
$\therefore y=4 x^{2}-14 x+12$
$=4 x-8 x-6 x+12$
$=4 x(x-2)-6(x-2)$
$=(x-2)(4 x-6)$
$y=4 x^{2}-14 x+q=0$
Since, $x=2$ is the root
$\therefore 4(2)^{2}-14(2)+q=0$
$\Rightarrow 16-28+q=0$
$\Rightarrow q=12$
$\therefore y=4 x^{2}-14 x+12$
$=4 x-8 x-6 x+12$
$=4 x(x-2)-6(x-2)$
$=(x-2)(4 x-6)$
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Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
