Solving quadratic equations 103
The quadratic equation x
2
+ 2x − 8 = 0 thus becomes
(x + 4)(x − 2) = 0
Since the only way that this can be true is for either the
first or the second or both factors to be zero,
either
(x + 4) = 0, i.e. x = −4
or
(x − 2) = 0, i.e. x = 2
Hence, the roots of x 2 + 2x − 8 = 0 are x = −4 and
x = 2.
Problem 3. Determine the roots of
x
2
− 6x + 9 = 0 by factorization
x 2 − 6x + 9 = (x − 3)(x − 3),
i.e.(x − 3) 2 = 0
The LHS is known as a perfect square.
Hence, x = 3 is the only root of the equation
x 2 − 6x + 9 = 0.
Problem 4. Solve the equation x 2 − 4x = 0
Factorizing gives x(x − 4) = 0
If
x(x − 4) = 0,
either
x = 0 or x − 4 = 0
i.e.
x = 0 or x = 4
These are the two roots of the given equation. Answers
can always be checked by substitution into the original
equation.
Problem 5. Solve the equation x 2 + 3x − 4 = 0
Factorizing gives
(x − 1)(x + 4) = 0
Hence, either
x − 1 = 0 or x + 4 = 0
i.e.
x = 1 or x = −4
Problem 6. Determine the roots of 4x 2 − 25 = 0
by factorization
The LHS of 4x 2 − 25 = 0 is the difference of two
squares, (2x) 2 and (5) 2 .
By factorizing, 4x 2 − 25 = (2x + 5)(2x − 5), i.e.
(2x + 5)(2x − 5) = 0
Hence, either
(2x + 5) = 0, i.e. x = −
5
2
= −2.5
or
(2x − 5) = 0, i.e. x =
5
2
= 2.5
Problem 7. Solve the equation x
2
− 5x + 6 = 0
Factorizing gives
(x − 3)(x − 2) = 0
Hence, either
x − 3 = 0 or x − 2 = 0
i.e.
x = 3 or x = 2
Problem 8. Solve the equation x 2 = 15 − 2x
Rearranging gives
x
2
+ 2x − 15 = 0
Factorizing gives
(x + 5)(x − 3) = 0
Hence, either
x + 5 = 0 or x − 3 = 0
i.e.
x = −5 or x = 3
Problem 9. Solve the equation 3x 2 − 11x − 4 = 0
by factorization
The factors of 3x 2 are 3x and x. These are placed in
brackets:
(3x
)(x
)
The factors of −4 are −4 and +1, or +4 and −1, or −2
and 2.
Remembering that the product of the two inner terms
added to the product of the two outer terms must equal
−11x, the only combination to give this is +1 and −4,
i.e.
3x
2
−11x − 4 = (3x + 1)(x − 4)
The quadratic equation 3x 2 − 11x − 4 = 0 thus
becomes
(3x + 1)(x − 4) = 0
Hence, either (3x + 1) = 0, i.e. x = −
1
3
or
(x − 4) = 0, i.e. x = 4
and both solutions may be checked in the original
equation.
Problem 10. Solve the quadratic equation
4x
2
+ 8x + 3 = 0 by factorizing
The factors of 4x 2 are 4x and x or 2x and 2x.
The factors of 3 are 3 and 1, or −3 and −1.
Remembering that the product of the inner terms added
to the product of the two outer terms must equal +8x,
the only combination that is true (by trial and error) is
(4x
2
+ 8x + 3) = (2x + 3)(2x + 1)
The quadratic equation x
2
+ 2x − 8 = 0 thus becomes
(x + 4)(x − 2) = 0
Since the only way that this can be true is for either the
first or the second or both factors to be zero,
either
(x + 4) = 0, i.e. x = −4
or
(x − 2) = 0, i.e. x = 2
Hence, the roots of x 2 + 2x − 8 = 0 are x = −4 and
x = 2.
Problem 3. Determine the roots of
x
2
− 6x + 9 = 0 by factorization
x 2 − 6x + 9 = (x − 3)(x − 3),
i.e.(x − 3) 2 = 0
The LHS is known as a perfect square.
Hence, x = 3 is the only root of the equation
x 2 − 6x + 9 = 0.
Problem 4. Solve the equation x 2 − 4x = 0
Factorizing gives x(x − 4) = 0
If
x(x − 4) = 0,
either
x = 0 or x − 4 = 0
i.e.
x = 0 or x = 4
These are the two roots of the given equation. Answers
can always be checked by substitution into the original
equation.
Problem 5. Solve the equation x 2 + 3x − 4 = 0
Factorizing gives
(x − 1)(x + 4) = 0
Hence, either
x − 1 = 0 or x + 4 = 0
i.e.
x = 1 or x = −4
Problem 6. Determine the roots of 4x 2 − 25 = 0
by factorization
The LHS of 4x 2 − 25 = 0 is the difference of two
squares, (2x) 2 and (5) 2 .
By factorizing, 4x 2 − 25 = (2x + 5)(2x − 5), i.e.
(2x + 5)(2x − 5) = 0
Hence, either
(2x + 5) = 0, i.e. x = −
5
2
= −2.5
or
(2x − 5) = 0, i.e. x =
5
2
= 2.5
Problem 7. Solve the equation x
2
− 5x + 6 = 0
Factorizing gives
(x − 3)(x − 2) = 0
Hence, either
x − 3 = 0 or x − 2 = 0
i.e.
x = 3 or x = 2
Problem 8. Solve the equation x 2 = 15 − 2x
Rearranging gives
x
2
+ 2x − 15 = 0
Factorizing gives
(x + 5)(x − 3) = 0
Hence, either
x + 5 = 0 or x − 3 = 0
i.e.
x = −5 or x = 3
Problem 9. Solve the equation 3x 2 − 11x − 4 = 0
by factorization
The factors of 3x 2 are 3x and x. These are placed in
brackets:
(3x
)(x
)
The factors of −4 are −4 and +1, or +4 and −1, or −2
and 2.
Remembering that the product of the two inner terms
added to the product of the two outer terms must equal
−11x, the only combination to give this is +1 and −4,
i.e.
3x
2
−11x − 4 = (3x + 1)(x − 4)
The quadratic equation 3x 2 − 11x − 4 = 0 thus
becomes
(3x + 1)(x − 4) = 0
Hence, either (3x + 1) = 0, i.e. x = −
1
3
or
(x − 4) = 0, i.e. x = 4
and both solutions may be checked in the original
equation.
Problem 10. Solve the quadratic equation
4x
2
+ 8x + 3 = 0 by factorizing
The factors of 4x 2 are 4x and x or 2x and 2x.
The factors of 3 are 3 and 1, or −3 and −1.
Remembering that the product of the inner terms added
to the product of the two outer terms must equal +8x,
the only combination that is true (by trial and error) is
(4x
2
+ 8x + 3) = (2x + 3)(2x + 1)