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Find the value of k for which one root of the quadratic equation kx² - 14 x 8 is equals to zero is 6 times the other?
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Find the value of k for which one root of the quadratic equation kx² -...
Introduction:
We are given a quadratic equation in the form of kx² - 14x + 8 = 0, where k is a constant. We need to find the value of k for which one root of the equation is 6 times the other.

Quadratic Equation:
A quadratic equation is of the form ax² + bx + c = 0, where a, b, and c are constants. In this case, the given equation is kx² - 14x + 8 = 0.

Roots of a Quadratic Equation:
The roots of a quadratic equation are the values of x for which the equation becomes zero. A quadratic equation can have either two distinct real roots, two complex roots, or one repeated root.

Let's Solve the Equation:
To find the value of k for which one root is 6 times the other, let's assume the roots of the equation to be p and 6p.

Using the sum and product of roots formula, we can write:
p + 6p = 14/ k (Sum of roots)
p * 6p = 8 / k (Product of roots)

Simplifying the above equations, we get:
7p = 14/ k
6p² = 8 / k

Substituting the Value of p:
Since we have the value of p in terms of k, we can substitute the value of p in the second equation to obtain a new equation in terms of k.

Substituting 7p = 14/ k in 6p² = 8 / k, we get:
6*(14/ k)² = 8 / k

Simplifying the equation further, we have:
6*(196/ k²) = 8 / k

Cross Multiplication:
To eliminate the fractions, we can cross multiply the equation:
6 * 196 = 8 * (k²)

Simplifying the equation, we get:
1176 = 8k²

Solving for k:
To find the value of k, we need to isolate k in the equation. Let's divide both sides of the equation by 8:
1176 / 8 = k²

Simplifying further, we have:
147 = k²

Taking the square root of both sides, we get:
k = ±√147

Therefore, the value of k for which one root of the quadratic equation kx² - 14x + 8 = 0 is 6 times the other is given by k = ±√147.
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Find the value of k for which one root of the quadratic equation kx² - 14 x 8 is equals to zero is 6 times the other?
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