Question

find the possible values of t if the roots of equation x2+t²= 6t+8x are real​

Answer 1

Formula: If $ax^{2}+bx+c=0$ be a quadratic equation with real roots, then

$\quad\quad \color{blue}{b^{2}-4ac\geq 0}$

Solution:

The given equation is

$\quad x^{2}+t^{2}=6t+8x$

$\Rightarrow x^{2}-8x+t^{2}-6t=0$

Comparing with the general form, we get

$\quad a=1,\:b=-8,\:c=t^{2}-6t$

For real roots, we must have

$\quad b^{2}-4ac\geq 0$

$\Rightarrow (-8)^{2}-4*1*(t^{2}-6t)\geq 0$

$\Rightarrow 64-4t^{2}+24t\geq 0$

$\Rightarrow t^{2}-6t-16\leq 0$

$\Rightarrow (t-8)(t+2)\leq 0$

$\Rightarrow t-8\leq 0\quad,\quad t+2\geq 0$

$\Rightarrow t\leq 8\quad,\quad t\geq -2$

$\Rightarrow -2\:\leq\:t\quad,\quad t\:\leq\:8$

$\Rightarrow -2\:\leq\:t\:\leq\:8$

Answer: $\color{blue}-2\:\leq\:t\:\leq\:8$

Answer 2

Given:

x2+t²= 6t+8x

To find:

find the possible values of t if the roots of equation x2+t²= 6t+8x are real​

Solution:

From given, we have a quadratic equation,

x² + t² = 6t + 8x

rearranging the terms we get,

x² - 8x - 6t  + t² = 0

we have, a = 1, b = - 8, c = - 6t  + t²

The condition for a quadratic equation to have real and unequal roots is given by, b² - 4ac ≥ 0

The condition for a quadratic equation to have real and equal roots is given by, b² - 4ac = 0

The condition for a quadratic equation to have real roots is given by,

b² - 4ac ≥ 0

(-8) ² - 4 × 1 × ( - 6t  + t² ) ≥ 0

64 + 24t - 4t² ≥ 0

solving the above quadratic equation, we get,

- (t + 2) × (t - 8) ≥ 0

⇒ (t + 2) × (t - 8) ≤ 0

t = -2 or -2 < t < 8 or t = 8

-2 ≤ t ≤ 8 is the root of given equation.