After substitution, you should have a quadratic equation in $x$. Two distinct real solutions correspond to a positive discriminant.

To force both $x$-values to be negative, use the sum and product of the roots (Vieta's formulas) instead of solving for $x$ explicitly. © ‌a‌n⁠iкo.​ai

Enter the equations:

• $y=x+1$
• $y=a(x-2)^2$

Create a slider for $a$.

Adjust $a$ until the graphs intersect at two points (you should see two intersection markers). If there is one marker, the graphs are tangent; if none, there are no real solutions.

Click each intersection point and look at its coordinates. Adjust $a$ so that both intersection points have $x<0$.

Once you find a value of $a$ that gives two intersections with both $x$-values negative, compare that value (or its approximate range) to the answer choices to see which one fits. © а‌n‌iкο‌.аi

Set the two expressions for $y$ equal:

Expand $(x-2)^2=x^2-4x+4$ and move everything to one side:

A quadratic has two distinct real solutions when its discriminant is positive.

For $ax^2+(-4a-1)x+(4a-1)=0$,

So two distinct real solutions require $12a+1>0$, meaning $a>-\frac{1}{12}$ (and also $a\ne 0$ so it is truly quadratic).

For a quadratic with real roots, both roots are negative when:

• the sum of the roots is negative, and
• the product of the roots is positive.

Using Vieta's formulas for $ax^2+(-4a-1)x+(4a-1)=0$:

To make the sum negative, $4+\frac{1}{a}<0$. This forces $a<0$, and multiplying by negative $a$ flips the inequality:

If $a<0$, then $-\frac{1}{a}>0$, so $4-\frac{1}{a}=4+\left(-\frac{1}{a}\right)>0$, meaning the product condition is automatically satisfied.

Combine the requirements:

• Two distinct real solutions: $a>-\frac{1}{12}$
• Both $x$-coordinates negative: $-\frac{1}{4}<a<0$ А‌nікo Questiоn Bank

Together, this means:

Among the choices, only $-\frac{1}{24}$ lies in this interval, so the value of $a$ could be $-\frac{1}{24}$.