Solve quadratic equations with calculations Algebra I

The solutions of the equation ax^{2}+bx+c=0 are:

• x_{1}=\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}
• x_{2}=\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}

If b^{2}-4ac (which is called the discriminant) is negative then the equation has no real solutions.

If b^{2}-4ac is zero then the equation has one double solution.

To solve the problem, first write the given equation in standard form:

2x^{2}-3x-2=0

Applying the formulas:

x_{1}=\dfrac{-\left(-3\right)-\sqrt{\left(-3\right)^{2}-4\left(2\right)\left(-2\right)}}{2\left(2\right)}=\dfrac{3-\sqrt{25}}{4}=-\dfrac{1}{2}

x_{2}=\dfrac{-\left(-3\right)+\sqrt{\left(-3\right)^{2}-4\left(2\right)\left(-2\right)}}{2\left(2\right)}=\dfrac{3+\sqrt{25}}{4}=2

The solutions of the equation ax^{2}+bx+c=0 are:

• x_{1}=\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}
• x_{2}=\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}

If b^{2}-4ac (which is called the discriminant) is negative then the equation has no real solutions.

If b^{2}-4ac is zero then the equation has one double solution.

To solve the problem, first write the given equation in standard form:

4x^{2}-12x+9=0

Applying the formulas:

x_{1}=\dfrac{-\left(-12\right)-\sqrt{\left(-12\right)^{2}-4\left(4\right)\left(9\right)}}{2\left(4\right)}=\dfrac{12-\sqrt{0}}{8}=\dfrac{3}{2}

x_{2}=\dfrac{-\left(-12\right)+\sqrt{\left(-12\right)^{2}-4\left(4\right)\left(9\right)}}{2\left(4\right)}=\dfrac{12+\sqrt{0}}{8}=\dfrac{3}{2}

The solutions of the equation ax^{2}+bx+c=0 are:

• x_{1}=\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}
• x_{2}=\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}

If b^{2}-4ac (which is called the discriminant) is negative then the equation has no real solutions.

If b^{2}-4ac is zero then the equation has one double solution.

To solve the problem, first write the given equation in standard form:

6x^{2}+11x-35=0

Applying the formulas:

x_{1}=\dfrac{-11-\sqrt{\left(11\right)^{2}-4\left(6\right)\left(-35\right)}}{2\left(6\right)}=\dfrac{-11-\sqrt{961}}{12}=-\dfrac{7}{2}

x_{2}=\dfrac{-11+\sqrt{\left(11\right)^{2}-4\left(6\right)\left(-35\right)}}{2\left(6\right)}=\dfrac{-11+\sqrt{961}}{12}=\dfrac{5}{3}

The solutions of the equation ax^{2}+bx+c=0 are:

• x_{1}=\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}
• x_{2}=\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}

If b^{2}-4ac (which is called the discriminant) is negative then the equation has no real solutions.

If b^{2}-4ac is zero then the equation has one double solution.

To solve the problem, first write the given equation in standard form:

2x^{2}+15x+7=10x-3

2x^{2}+5x+10=0

Applying the formulas:

x_{1}=\dfrac{-5-\sqrt{\left(5\right)^{2}-4\left(2\right)\left(10\right)}}{2\left(2\right)}=\dfrac{-5-\sqrt{-55}}{4} which is not a real number.

x_{2}=\dfrac{-5+\sqrt{\left(5\right)^{2}-4\left(2\right)\left(10\right)}}{2\left(2\right)}=\dfrac{-5+\sqrt{-55}}{4} which is not a real number.

The solutions of the equation ax^{2}+bx+c=0 are:

• x_{1}=\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}
• x_{2}=\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}

If b^{2}-4ac (which is called the discriminant) is negative then the equation has no real solutions.

If b^{2}-4ac is zero then the equation has one double solution.

To solve the problem, first write the given equation in standard form:

x^{2}-16x=0

Applying the formulas:

x_{1}=\dfrac{-\left(-16\right)-\sqrt{\left(-16\right)^{2}-4\left(1\right)\left(0\right)}}{2\left(1\right)}=\dfrac{16-\sqrt{256}}{2}=0

x_{2}=\dfrac{-\left(-16\right)+\sqrt{\left(-16\right)^{2}-4\left(1\right)\left(0\right)}}{2\left(1\right)}=\dfrac{16+\sqrt{256}}{2}=16

The solutions of the equation ax^{2}+bx+c=0 are:

• x_{1}=\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}
• x_{2}=\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}

If b^{2}-4ac (which is called the discriminant) is negative then the equation has no real solutions.

If b^{2}-4ac is zero then the equation has one double solution.

To solve the problem, first write the given equation in standard form:

2x^{2}-x-6=0

Applying the formulas:

x_{1}=\dfrac{-\left(-1\right)-\sqrt{\left(-1\right)^{2}-4\left(2\right)\left(-6\right)}}{2\left(2\right)}=\dfrac{1-\sqrt{49}}{4}=-\dfrac{3}{2}

x_{2}=\dfrac{-\left(-1\right)+\sqrt{\left(-1\right)^{2}-4\left(2\right)\left(-6\right)}}{2\left(2\right)}=\dfrac{1+\sqrt{49}}{4}=2

The solutions of the equation ax^{2}+bx+c=0 are:

• x_{1}=\dfrac{-b-\sqrt{b^{2}-4ac}}{2a}
• x_{2}=\dfrac{-b+\sqrt{b^{2}-4ac}}{2a}

If b^{2}-4ac (which is called the discriminant) is negative then the equation has no real solutions.

If b^{2}-4ac is zero then the equation has one double solution.

To solve the problem, first write the given equation in standard form:

3x^{2}-14x-5=-x^{2}-12x-4

4x^{2}-2x-1=0

Applying the formulas:

x_{1}=\dfrac{-\left(-2\right)-\sqrt{\left(-2\right)^{2}-4\left(4\right)\left(-1\right)}}{2\left(4\right)}=\dfrac{2-\sqrt{20}}{8}=\dfrac{2-2\sqrt{5}}{8}=\dfrac{1-\sqrt{5}}{4}

x_{2}=\dfrac{-\left(-2\right)+\sqrt{\left(-2\right)^{2}-4\left(4\right)\left(-1\right)}}{2\left(4\right)}=\dfrac{2+\sqrt{20}}{8}=\dfrac{2+2\sqrt{5}}{8}=\dfrac{1+\sqrt{5}}{4}