Mathematics 10 — Exercise 2.1 · Q2 (iii)

Full question

Solve the following quadratic equation by factorization method: 6 × x²+13 × x-5 = 0.

Step-by-step solution

1. Substitute the function expressions, keeping the brackets.

   $$6\times x^2+13\times x-5-0$$

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

   $$6\times x^2+13\times x-5$$

3. Write EVERY distributed product before collecting terms.

   $$-5+6\times x^2+13\times x-0$$

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

   $$-5=-5$$

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

   $$6\times x^2=6\times x^2$$

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

   $$13\times x=13\times x$$

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

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

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

   $$-5+0=-5$$

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

   $$6=6$$

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

    $$13=13$$

11. Put the simplified groups together.

    $$6\times x^2+13\times x-5$$

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

    $$6\times x^2+13\times x-5$$

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

    $$6\times x^2+13\times x-5=0$$

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

    $$(-2+15=13,\left(-2\right)\times 15=-30)$$

15. Rewrite the middle term using those two parts.

    $$6\times x^2-2\times x+15\times x-5=0$$

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

    $$\left(2\times x\right)\times \left(3\times x-1\right)+5\times \left(3\times x-1\right)=0$$

17. The common bracket can now be factored out.

    $$\left(2\times x+5\right)\times \left(3\times x-1\right)=0$$

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

    $$(2\times x+5=0,3\times x-1=0)$$

19. Take this factor equal to zero.

    $$2\times x+5=0$$

20. Subtract the constant from both sides.

    $$2\times x+5-5=0-5$$

21. Simplify both sides.

    $$2\times x=-5$$

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

    $$x=-\frac{5}{2}$$

23. Reduce the fraction to obtain this root.

    $$x=-\frac{5}{2}$$

24. Take this factor equal to zero.

    $$3\times x-1=0$$

25. Subtract the constant from both sides.

    $$3\times x-1+1=0+1$$

26. Simplify both sides.

    $$3\times x=1$$

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

    $$x=\frac{1}{3}$$

28. Reduce the fraction to obtain this root.

    $$x=\frac{1}{3}$$

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

    $$0=0$$

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

    $$0=0$$

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

    $$0=0$$

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

    $$0=0$$

Answer

The final answer is shown below.