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Solving Simple Quadratic Equations

Algebra and Graphs Solving Simple Quadratic Equations

Courses Overview

🎬 Video Tutorial

(0:01) Solving Simple Quadratic Equations: For $x^2 = k$, there are two solutions: $x = \pm \sqrt{k}$. Note that $k$ must be non-negative, as $x^2$ cannot be negative.
(1:26) No Real Solutions for Negative Numbers: If $k$ is negative, such as in $x^2 = -25$, there are no real solutions.
(1:55) Solving $x^2 – d x = 0$ by Factorisation: Factor out $x$ to get $x(x – d) = 0$, giving solutions $x = 0$ and $x = d$.
(2:45) A Common Pitfall: Avoid dividing both sides by $x$, as it can lead to missing solutions, especially $x = 0$.

📂 Revision Cards

Solving the quadratic equation x² = k, showing the general solution x equals plus or minus the square root of k, and a specific example x² = 9 with solutions. — 1) Solving Simple Quadratics

Solving quadratic equations with the steps to isolate x² and find solutions x sub one equals the positive square root of k. — 2) Solving simple quadratic equations

Solving quadratic equations, demonstrating x² = k and x² + 50 = 0 leading to no real solution since x² = -25. — 3) Solving simple quadratic equations

Solving quadratic equations by factorisation, showing x² - dx = 0 with solutions x sub one equals zero, and x sub two equals d. — 4) Solving simple quadratic equations

Solving quadratic equation x² - dx = 0 with steps to simplify and factorise, showing solutions x? = 0 and x? = 5. — 5) Solving simple quadratic equations

Solving simple quadratic equations with examples: x squared equals k, leading to x equals plus or minus the square root of k, and x² - dx = 0 leading to x = 0 or x = d. — 6) Overview Solving Simple Quadratic Equations

🍪 Quiz Time - Practice Now!

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Solving Simple Quadratic Equations

1 / 6

Q: What's the solution for $x^2 = 25$?

$x = 25, x = -25$

$x = -5$

$x = 5, x = -5$

$x = 5$

Feedback: Not quite! To solve $x^2 = 25$, take the square root of both sides. Since both $5$ and $-5$ squared give 25, the solutions are $x = 5$ and $x = -5$.

Feedback: Well done! To solve $x^2 = 25$, take the square root of both sides. Since both $5$ and $-5$ squared give 25, the solutions are $x = 5$ and $x = -5$.

2 / 6

Q: What's the solution for $x^2 = 0$?

$x = 1$

$x = 0$

$x = 0, x = 1$

No real solutions

Feedback: Not quite! The only solution is $x = 0$, because $0^2 = 0$. $1^2 = 1$, so $x = 1$ is not a solution.

Feedback: Well done! The only solution is $x = 0$, because $0^2 = 0$. $1^2 = 1$, so $x = 1$ is not a solution.

3 / 6

Q: What's the solution for $x^2 = -16$?

No real solutions

$x = 4, x = -4$

$x = -4$

$x = 4$

Feedback: Not quite! Since $x^2 = -16$ has a negative value on the right side, there are no real solutions. This is because the square of any real number is always positive or zero.

Feedback: Well done! Since $x^2 = -16$ has a negative value on the right side, there are no real solutions. This is because the square of any real number is always positive or zero.

4 / 6

Q: What's the solution for $x^2 - 64 = 0$?

$x = 6, x = -6$

$x = 8$

$x = 8, x = -8$

$x = 4$

Feedback: Not quite! To solve $x^2 - 64 = 0$, first isolate $x^2$ to get $x^2 = 64$. Taking the square root of both sides, we find that $\lvert x \lvert = 8$. Therefore, the solutions are $x = 8$ and $x = -8$.

Feedback: Well done! To solve $x^2 - 64 = 0$, first isolate $x^2$ to get $x^2 = 64$. Taking the square root of both sides, we find that $\lvert x \lvert = 8$. Therefore, the solutions are $x = 8$ and $x = -8$.

5 / 6

Q: What's the solution for $x^2 - 81 = 0$?

$x = 9$

$x = 8$

$x = 8, x = -8$

$x = 9, x = -9$

Feedback: Not quite! To solve $x^2 - 81 = 0$, first isolate $x^2$ to get $x^2 = 81$. Taking the square root of both sides, we find that $\lvert x \rvert = 9$. Therefore, the solutions are $x = 9$ and $x = -9$.

Feedback: Well done! To solve $x^2 - 81 = 0$, first isolate $x^2$ to get $x^2 = 81$. Taking the square root of both sides, we find that $\lvert x \rvert = 9$. Therefore, the solutions are $x = 9$ and $x = -9$.

6 / 6

Q: What is the solution for $5x^2 - 125 = 0$?

$x = 25$

$x = -5, x = 5$

$x = -25, x = 25$

$x = 5$

Feedback: Not quite! First isolate $x^2$ by adding 125 to both sides: $5x^2 = 125$. Then divide both sides by 5 to get $x^2 = 25$. Taking the square root of both sides, we get $\lvert x \rvert = 5$. Therefore, the solutions are $x = 5$ and $x = -5$.

Feedback: Well done! First isolate $x^2$ by adding 125 to both sides: $5x^2 = 125$. Then divide both sides by 5 to get $x^2 = 25$. Taking the square root of both sides, we get $\lvert x \rvert = 5$. Therefore, the solutions are $x = 5$ and $x = -5$.

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