Mathematics 10 — Exercise 5.4 · Q1 (iv)

Full question

Solve the following equation by reducing it to a quadratic equation. Check the resulting values in the original equation.

Step-by-step solution

1. Use the exponent laws to write the equation as a quadratic in t, where t is the following positive power.

   $$t=2^x$$

2. After substitution the quadratic equation is:

   $$8\times t^2-9\times t+1=0$$

3. Substitute the function expressions, keeping the brackets.

   $$8\times t^2-9\times t+1-0$$

4. Remove the brackets. A preceding minus sign reverses every sign inside that bracket.

   $$8\times t^2-9\times t+1$$

5. Write EVERY distributed product before collecting terms.

   $$1-9\times t+8\times t^2-0$$

6. Calculate this individual product, including its sign and powers.

   $$1=1$$

7. Calculate this individual product, including its sign and powers.

   $$-9\times t=-9\times t$$

8. Calculate this individual product, including its sign and powers.

   $$8\times t^2=8\times t^2$$

9. Calculate this individual product, including its sign and powers.

   $$\left(-1\right)\times 0=0$$

10. These terms have the same variable part. Add ONLY their coefficients.

    $$1+0=1$$

11. These terms have the same variable part. Add ONLY their coefficients.

    $$-9=-9$$

12. These terms have the same variable part. Add ONLY their coefficients.

    $$8=8$$

13. Put the simplified groups together.

    $$8\times t^2-9\times t+1$$

14. Collect terms with the same power and simplify their coefficients.

    $$8\times t^2-9\times t+1$$

15. Expand the brackets and subtract the right-hand terms from both sides. Collect powers of the variable.

    $$8\times t^2-9\times t+1=0$$

16. Split the middle coefficient into two numbers whose sum is b and whose product is a·c.

    $$(-1-8=-9,\left(-1\right)\times \left(-8\right)=8)$$

17. Rewrite the middle term using those two parts.

    $$8\times t^2-t-8\times t+1=0$$

18. Group the first two terms and the last two. Take out the common factor from each group.

    $$t\times \left(8\times t-1\right)-(8\times t-1)=0$$

19. The common bracket can now be factored out.

    $$\left(t-1\right)\times \left(8\times t-1\right)=0$$

20. A product is zero only if at least one factor is zero. Set each linear factor equal to zero.

    $$(t-1=0,8\times t-1=0)$$

21. Take this factor equal to zero.

    $$t-1=0$$

22. Subtract the constant from both sides.

    $$t-1+1=0+1$$

23. Simplify both sides.

    $$t=1$$

24. Divide both sides by the coefficient of the variable.

    $$t=\frac{1}{1}$$

25. Reduce the fraction to obtain this root.

    $$t=1$$

26. Take this factor equal to zero.

    $$8\times t-1=0$$

27. Subtract the constant from both sides.

    $$8\times t-1+1=0+1$$

28. Simplify both sides.

    $$8\times t=1$$

29. Divide both sides by the coefficient of the variable.

    $$t=\frac{1}{8}$$

30. Reduce the fraction to obtain this root.

    $$t=\frac{1}{8}$$

31. Substitute this root into the original equation before simplifying.

    $$0=0$$

32. Calculate both sides; their equal values confirm the root.

    $$0=0$$

33. Substitute this root into the original equation before simplifying.

    $$0=0$$

34. Calculate both sides; their equal values confirm the root.

    $$0=0$$

35. Set the original power equal to this positive t value.

    $$2^x=\frac{1}{8}$$

36. Write both sides with the same base if possible; otherwise use a logarithm.

    $$x=-\frac{\log \left(8\right)}{\log \left(2\right)}$$

37. Divide the exponent equation by the multiplier of x.

    $$x=-3$$

38. Set the original power equal to this positive t value.

    $$2^x=1$$

39. Write both sides with the same base if possible; otherwise use a logarithm.

    $$x=0$$

40. Divide the exponent equation by the multiplier of x.

    $$x=0$$

41. Collect every valid root from every substitution branch.

    $$\{-3,0\}$$

Answer

The final answer is shown below.