1.RESOLVENDO EQUACÓES DO 2^circ GRAIJ Exercício 1. Resolva as seguintes equações , cfetuando a prova real ao final: (a) x^2-6x+8=0 x^2=25 x^2+2x+1=0 2x^2-8x-42=0 (c) -x^2-10x-16=0 x^2+x+28=0 x^2-6x+9=0 (f) x^2-11x+18=0 (h) (i) 2x^2+18x+40=0 (j) x^2-5x=0 (k) 3x^2=-3x (I) x^2+3=0 (m) -6x-9x^2-1=0 ) x^2+2x+2=0 -x^2+8x=0 (p) x^2-64=0 (q) -x^2+5x-4=0 (r) -6x+x^2=0 -3x+2=-2x^2 (t) 30x^2=14x+4 (u) x^2+2x=80 (y) 12x^2-7x=-1 (w)x^2=1—— x (x) -12x^2+13x=3 21x^2+34x+8=0

Para resolver as equações do segundo grau, podemos utilizar a fórmula de Bhaskara:<br /><br />\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]<br /><br />Onde a, b e c são os coeficientes da equação.<br /><br />Vamos resolver algumas das equações fornecidas:<br /><br />(a) \( x^2 - 6x + 8 = 0 \)<br /><br />Aplicando a fórmula de Bhaskara:<br /><br />\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(8)}}{2(1)} \]<br />\[ x = \frac{6 \pm \sqrt{36 - 32}}{2} \]<br />\[ x = \frac{6 \pm \sqrt{4}}{2} \]<br />\[ x = \frac{6 \pm 2}{2} \]<br />\[ x = 3 \pm 1 \]<br />\[ x = 2 \quad \text{ou} \quad x = 4 \]<br /><br />(b) \( x^2 = 25 \)<br /><br />\[ x = \pm \sqrt{25} \]<br />\[ x = \pm 5 \]<br /><br />(c) \( x^2 + 2x + 1 = 0 \)<br /><br />Aplicando a fórmula de Bhaskara:<br /><br />\[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(1)}}{2(1)} \]<br />\[ x = \frac{-2 \pm \sqrt{4 - 4}}{2} \]<br />\[ x = \frac{-2 \pm 0}{2} \]<br />\[ x = -1 \]<br /><br />(d) \( 2x^2 - 8x - 42 = 0 \)<br /><br />Aplicando a fórmula de Bhaskara:<br /><br />\[ x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(2)(-42)}}{2(2)} \]<br />\[ x = \frac{8 \pm \sqrt{64 + 336}}{4} \]<br />\[ x = \frac{8 \pm \sqrt{400}}{4} \]<br />\[ x = \frac{8 \pm 20}{4} \]<br />\[ x = 5 \quad \text{ou} \quad x = -3 \]<br /><br />(e) \( -x^2 - 10x - 16 = 0 \)<br /><br />Multiplicando por -1:<br /><br />\[ x^2 + 10x + 16 = 0 \]<br /><br />Aplicando a fórmula de Bhaskara:<br /><br />\[ x = \frac{-10 \pm \sqrt{10^2 - 4(1)(16)}}{2(1)} \]<br />\[ x = \frac{-10 \pm \sqrt{100 - 64}}{2} \]<br />\[ x = \frac{-10 \pm \sqrt{36}}{2} \]<br />\[ x = \frac{-10 \pm 6}{2} \]<br />\[ x = -5 \pm 3 \]<br />\[ x = -2 \quad \text{ou} \quad x = -8 \]<br /><br />(g) \( x^2 - 6x + 9 = 0 \)<br /><br />\[ (x - 3)^2 = 0 \]<br />\[ x = 3 \]<br /><br />(h) \( x^2 - 11x + 18 = 0 \)<br /><br />Aplicando a fórmula de Bhaskara:<br /><br />\[ x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(1)(18)}}{2(1)} \]<br />\[ x = \frac{11 \pm \sqrt{121 - 72}}{2} \]<br />\[ x = \frac{11 \pm \sqrt{49}}{2} \]<br />\[ x = \frac{11 \pm 7}{2} \]<br />\[ x = 9 \quad \text{ou} \quad x = 2 \]<br /><br />(i) \( 2x^2 + 18x + 40 = 0 \)<br /><br />Aplicando a fórmula de Bhaskara:<br /><br />\[ x = \frac{-18 \pm \sqrt{18^2 - 4(2)(40)}}{2(2)} \]<br />\[ x = \frac{-18 \pm \sqrt{324 - 320}}{4} \]<br />\[ x = \frac{-18 \pm \sqrt{4}}{4} \]<br />\[ x = \frac{-18 \pm 2}{4} \]<br />\[ x = -4.5 \pm 0.5 \]<br />\[ x = -5 \quad \text{ou} \quad x = -4 \]<br /><br />(j) \( x^2 -