Solving quadratics

For any quadratic written as $a{x}^2+bx+c=0$, the discriminant $\Delta$ is the value used to study the roots (this appears in Functions and Quadratic Relationships).

The discriminant is defined by: $\Delta =b^2-4ac$

We are told that the discriminant is not just any value, but exactly $b^2$. So we set the usual discriminant expression equal to $b^2$:

$b^2-4ac=b^2$

Now solve algebraically for $c$:

$\begin{array}{rl}{b}^2-4ac& =b^2\\ -4ac& =b^2-b^2\\ -4ac& =0\end{array}$

Because it is a quadratic, $a\ne 0$. Dividing both sides by $-4a$ gives:

$c=0$

Putting $c=0$ into $a{x}^2+bx+c=0$ leaves:

$a{x}^2+bx=0$

Factor out $x$:

$x(ax+b)=0$

So the roots are: $x=0\text{or}x=-\frac{b}{a}$

• Option B: A repeated root happens when $\Delta =0$. Here $\Delta =b^2$, which is only $0$ if $b=0$, so this is not guaranteed.
• Option C: Since $b^2\ge 0$, we have $\Delta \ge 0$, so real solutions must exist.
• Option D: The roots are $0$ and $-b/a$, which are not necessarily $b$ and $-b$ unless special values of $a$ and $b$ are chosen.

From $\Delta =b^2$ we deduced $c=0$, so the equation becomes $x(ax+b)=0$. That form always includes $x=0$ as a solution, so the necessary conclusion is that one of the solutions must be 0.

The problem gives a special situation for $a{x}^2+bx+c=0$: the discriminant is exactly $b^2$.

$\Delta =b^2$

Any quadratic equation can be solved using the quadratic formula (listed in Topic 2: Functions):

$\begin{array}{rl}x& =\frac{-b\pm \sqrt{\Delta }}{2a}\end{array}$

Replace $\Delta$ with $b^2$:

$\begin{array}{rl}x& =\frac{-b\pm \sqrt{b^2}}{2a}\\ x& =\frac{-b\pm b}{2a}\end{array}$

Now evaluate each sign choice.

Case 1 (Addition): $\begin{array}{rl}{x}_1& =\frac{-b+b}{2a}\\ x_1& =\frac{0}{2a}=0\end{array}$

Case 2 (Subtraction): $\begin{array}{rl}{x}_2& =\frac{-b-b}{2a}\\ x_2& =\frac{-2b}{2a}=-\frac{b}{a}\end{array}$

Because $x_1$ always simplifies to $0$, one solution must be zero.

• Option B: A double root requires $\Delta =0$. Here $\Delta =b^2$, which only equals $0$ when $b=0$.
• Option C: No real roots would require $\Delta <0$, but $b^2$ is always $\ge 0$ for real $b$.
• Option D: The roots found are $0$ and $-b/a$, not $b$ and $-b$.

The formula gives one root as $0$ every time, so the statement that fits is Option A: One of the solutions must be 0.

To test the condition by example, choose convenient values for $a$ and $b$, then find $c$ so that $\Delta =b^2$.

Using the discriminant formula:

$\Delta =b^2-4ac$

Let $a=1$ and $b=2$. The condition $\Delta =b^2$ becomes:

$\begin{array}{rl}{b}^2-4ac& =b^2\\ 2^2-4(1)c& =2^2\\ 4-4c& =4\end{array}$

Solve for $c$:

$\begin{array}{rl}-4c& =0\\ c& =0\end{array}$

With $a=1$, $b=2$, and $c=0$, the equation is:

$1x^2+2x+0=0$ $x^2+2x=0$

Factor:

$x(x+2)=0$

So:

$\begin{array}{rl}x& =0\\ x& =-2\end{array}$

• For Option B: The roots are $0$ and $-2$, which are different, so there is not a repeated root.
• For Option C: Both solutions are real, so it is not true that there are no real solutions.
• For Option D: Here $b=2$, but the solutions are $0$ and $-2$, not $2$ and $-2$, so that claim is not always true.

In this example, one solution is $0$. More importantly, the condition $\Delta =b^2$ forces $c=0$, so the quadratic always has the form $a{x}^2+bx=0$. Factoring gives $x(ax+b)=0$, which guarantees $x=0$ is a root.

Therefore, the correct conclusion is Option A.

Before using the quadratic formula, the equation must match the standard quadratic form:

$a{x}^2+bx+c=0$

Here:

• $a$ is the number multiplying $x^2$
• $b$ is the number multiplying $x$
• $c$ is the constant term

Recognizing this form is a key skill in Topic 3: Functions and Quadratic Relationships.

The equation is $x^2-9=0$. To compare it directly with $a{x}^2+bx+c=0$, we make the missing parts explicit:

• The $x^2$ term has an implied coefficient of $1$.
• There is no $x$ term, so its coefficient is $0$ (written as $0x$).
• The constant term is $-9$.

So we can write: $1x^2+0x+(-9)=0$

Now compare $1x^2+0x+(-9)=0$ with $a{x}^2+bx+c=0$. This gives:

$\begin{array}{rl}a& =1\\ b& =0\\ c& =-9\end{array}$

So the coefficient choices stated in the question are exactly the ones needed for the quadratic formula.

Because the coefficients have been identified correctly, the statement cannot be wrong.

So we cross out the incorrect choice.

The claim is correct: $a=1$, $b=0$, and $c=-9$ are the right values to substitute into the quadratic formula.

To use the quadratic formula, the equation should be in the form $a{x}^2+bx+c=0$.

From $x^2-9=0$:

• The coefficient of $x^2$ is $1$, so $a=1$.
• The $x$ term is missing, so its coefficient is $0$, so $b=0$.
• The constant term is $-9$, so $c=-9$.

Use the quadratic formula:

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

Putting in $a=1$, $b=0$, and $c=-9$:

$x=\frac{-0\pm \sqrt{0^2-4(1)(-9)}}{2(1)}$

Now simplify step by step:

$\begin{array}{rl}x& =\frac{0\pm \sqrt{0-(-36)}}{2}\\ x& =\frac{\pm \sqrt{36}}{2}\\ x& =\frac{\pm 6}{2}\end{array}$

So the two answers are: $x=\frac{6}{2}=3\text{and}x=\frac{-6}{2}=-3$

Using the quadratic formula produced $x=3$ and $x=-3$, which are exactly the solutions to $x^2-9=0$. That means using $a=1$, $b=0$, and $c=-9$ is valid.

So the false option can be ruled out.

Since the substitution works and gives correct solutions, the statement must be true.