Question

4х²-8х+4=0 х²+6х-7=0 3х²+6х+3=0 х²+8х+16=0 18х²-9х=0

Answer 1

$4 {x}^{2} - 8x + 4 = 0 \div 4 \\ {x}^{2} - 2x + 1 = 0 \\ d = 4 - 4 = 0$

$x = \frac{2 + 0}{2} = \frac{2}{2} = 1$

${x}^{2} + 6x - 7 = 0 \\ d = 36 + 28 = 64 = 4 \\ x1 = \frac{ - 6 + 4}{2} = - \frac{2}{2} = - 1$

$x2 = \frac{ - 6 - 4}{2} = - \frac{10}{2} = - 5$

$3 {x}^{2} + 6x + 3 = 0 \div 3 \\ {x}^{2} + 2x + 1 = 0 \\ d = 4 - 4 = 0 \\ x = \frac{2 + 0}{2} = \frac{2}{2} = 1$

х=1

${x}^{2} + 8x + 16 = 0 \\ d = 64 - 64 = 0 \\ x = \frac{8 + 0}{2} = \frac{8}{2} = 4$

${18x}^{2} - 9x = 0 \\ 9x(2x - 1) = 0 \\ 9x = 0 \\ x = 0$

$2x - 1 = 0 \\ 2x = 1 \\ x = \frac{1}{2} = 0.5$

Answer 2

It looks like you've provided several quadratic equations. If you need help with solving them, I can certainly assist you. To solve a quadratic equation in the form $ax^2 + bx + c = 0$, you can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Let's go through each of the equations you provided and find their solutions.

1. $4x^2 - 8x + 4 = 0$ Here, $a = 4$, $b = -8$, and $c = 4$. Using the quadratic formula: $x = \frac{8 \pm \sqrt{(-8)^2 - 4(4)(4)}}{2(4)}$ $x = \frac{8 \pm 0}{8}$ $x = 1$

2. $x^2 + 6x - 7 = 0$ Here, $a = 1$, $b = 6$, and $c = -7$. Using the quadratic formula: $x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-7)}}{2(1)}$ $x = \frac{-6 \pm \sqrt{64}}{2}$ $x = \frac{-6 \pm 8}{2}$ So, $x = -7$ or $x = 1$.

3. $3x^2 + 6x + 3 = 0$ Here, $a = 3$, $b = 6$, and $c = 3$. Using the quadratic formula: $x = \frac{-6 \pm \sqrt{6^2 - 4(3)(3)}}{2(3)}$ $x = \frac{-6 \pm 0}{6}$ $x = -1$

4. $x^2 + 8x + 16 = 0$ Here, $a = 1$, $b = 8$, and $c = 16$. Using the quadratic formula: $x = \frac{-8 \pm \sqrt{8^2 - 4(1)(16)}}{2(1)}$ $x = \frac{-8 \pm 0}{2}$ $x = -4$

5. $18x^2 - 9x = 0$ Factor out $9x$ from the equation: $9x(2x - 1) = 0$ Setting each factor to zero gives $x = 0$ or $x = \frac{1}{2}$.

So, the solutions for the given equations are $x = 1$, $x = -7$ or $x = 1$, $x = -1$, $x = -4$, and $x = 0$ or $x = \frac{1}{2}$.