Mathematics - Quadratic Equation Question with Solution | TestHub

For a quadratic equation to have real roots, the discriminant must be greater than or equal to zero.

$$Disc\ge 0\Rightarrow (a-4{)}^2-4(a^2-3a+3)\ge 0$$

Expanding and simplifying the inequality:

$$a^2+16-8a-4a^2+12a-12\ge 0$$

$$-3a^2+4a+4\ge 0$$

Multiplying by -1 and reversing the inequality sign:

$$3a^2-4a-4\le 0$$

Factoring the quadratic expression:

$$3a^2-6a+2a-4\le 0$$

$$3a(a-2)+2(a-2)\le 0$$

$$(3a+2)(a-2)\le 0$$

This inequality holds true when $a$ is between $-\frac{2}{3}$ and $2$, inclusive:

$$-\frac{2}{3}\le a\le 2$$

Now, consider the condition ${\alpha }^2+{\beta }^2=6$. We know that ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$.

From Vieta's formulas, for a quadratic equation $A{x}^2+Bx+C=0$, the sum of the roots is $\alpha +\beta =-\frac{B}{A}$ and the product of the roots is $\alpha \beta =\frac{C}{A}$.

Assuming the original quadratic equation is $x^2-(a-4)x+(a^2-3a+3)=0$, then:

$$\alpha +\beta =-(a-4)=4-a$$

$$\alpha \beta =a^2-3a+3$$

Substituting these into the equation for ${\alpha }^2+{\beta }^2$:

$$(4-a{)}^2-2(a^2-3a+3)=6$$

Expanding and simplifying:

$$16-8a+a^2-2a^2+6a-6=6$$

$$-a^2-2a+10=6$$

$$-a^2-2a+4=0$$

Multiplying by -1:

$$a^2+2a-4=0$$

Using the quadratic formula $a=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$:

$$a=\frac{-2\pm \sqrt{2^2-4(1)(-4)}}{2(1)}$$

$$a=\frac{-2\pm \sqrt{4+16}}{2}$$

$$a=\frac{-2\pm \sqrt{20}}{2}$$

$$a=\frac{-2\pm 2\sqrt{5}}{2}$$

$$a=-1\pm \sqrt{5}$$