Mathematics - MCQ Practice Questions

Step 1: Apply Vieta's Formulas

For the quadratic $3x^2-4x+k=0$, Vieta's formulas give:

**Step 2: Expand ${\alpha }^2+{\beta }^2$

Using the identity ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$:

Step 3: Substitute the Given Value

Setting ${\alpha }^2+{\beta }^2=\frac{10}{9}$:

**Step 4: Solve for $k$

Multiplying both sides by $\frac{3}{2}$:

Step 1: Express terms using the common ratio

Since $a,b,c$ are in geometric progression with common ratio $r$,

Step 2: Set up the two equations

Substituting into the given conditions:

Step 3: Divide the second equation by the first

Squaring the first equation gives $a^2(1+r+r^2{)}^2=1521$. Dividing the second equation by this:

Step 4: Simplify using the factorization

Note that $1+r^2+r^4=(1+r+r^2)(1-r+r^2)$, so:

This simplifies to:

**Step 5: Solve for $r$

Cross-multiplying:

So $r=2$ or $r=\frac{1}{2}$.

**Step 6: Verify with $a=3$

Taking $r=2$: from $a(1+r+r^2)=39$ we get $a(1+2+4)=7a=39$... let us check $r=2$ with the ratio $39/7$. Indeed both values $r=2$ (with $a=3$, giving $3+6+12=21$...

Using $a=3,r=2$: $3+6+12=21\ne 39$. The value consistent with all conditions is: